59863727 worktext in differential equations

1

STANDARD INTEGRATION FORMULAS

(1) ∫ du=u+c

(2) ∫ adu=a∫ du

(3) ∫ (du+dv−dw )=∫ du+∫dv−∫ dw

(4) ∫undu= un+1

n+1+c ,(n≠−1)

(5) ∫u−1du=∫ duu

=lnu+c ,u>0

(6) ∫ audu= au

lna+c ,a>0

(7) ∫ eudu=eu+c , e=2.71828…

(8) ∫cos udu=sinu+c

(9) ∫sinudu=−cosu+c

(10) ∫ sec2u du=tan u+c

(11) ∫ csc2udu=−cotu+c

(12) ∫ sec u tan udu=sec u+c

(13) ∫ cscucotudu=−csc u+c

(14) ∫ tan udu=−lncos u+c=ln secu+c

(15) ∫cotu du=lnsinu+c=−ln csc u+c

(16) ∫ sec udu=ln ¿¿¿

(17) ∫ cscudu=ln ¿¿¿

2

(18) ∫ du

√a2−u2=arc sin

ua

+c

(19) ∫ du

u√u2+a2=−1

aln( a+√u2+a2

u )

(20) ∫ du

u√u2−a2= 1

aarc sec

ua

+c

(21) ∫ duu2−a2 = 1

2aln [ u−a

u+a ]+c

(22) ∫ dua2−u2 = 1

2aln [ u+a

u−a ]+c

(23) ∫ u

√u2±a2=ln (u+√u2±a2) + c

(24) ∫ du

a2+u2= 1

aarc tan

ua

+c

(25) ∫ du

(au2+bu+c )1 /2= 1

√aln {(2au+b )+2√a(au2+bu+c )1 /2 }+c

(26) ∫cosh udu=sinhu+c (27) ∫sinhudu=cosh u+c

(28) ∫ sech2udu=tanh u+c

(29) ∫ csch2udu=−cothu+c

(30) ∫sinhudu=cosh u+c

(31) ∫ sechu tanhudu=−sechu+c

(32) ∫ cschucoth udu=−cschu+c

(33) ∫udv=uv−∫vdu

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Exercise No. 1

Evaluate the following integrals:

1. ∫cos12ydy

2. ∫cos4 x sin x dx

3. ∫ sin t dt

(4+cos t )3 /2

4

4. ∫ tan2by dy

5. ∫sin2 x tan x dx

6. ∫ (3 x−2 ) dxx2+2x+17

7. ∫ x2 e−x dx

5

8. ∫ ysec 2 y dy

9. ∫ (4 x−1 )dx(2x+1 )3 /2

10. ∫ dx

x (x2−a2)

11. ∫ dx

1+ex

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12. ∫ dx

x2−5 x−2

13.(2x−3 )dx

( x−1 )2 ( x−2 )2

14. ∫ lnx dxx

15. ∫ ln ( lnx )dxxln x

7

16. ∫ ax ex dx

17. ∫ dx

√12x−4 x2−5

18. ∫ ln ( x2+a2 )dx

8

19. ∫ xlnx dx

20. ∫ dxx+3+√ x+3

Unit 1

DEFINITIONS AND NOTATIONS

1. A “differential equation” (abbreviated as D.E.) is one which contains at least one derivative. Most often, a D.E. is written in terms of differential – a more convenient form for analysis and solutions.

2. When an equation involves one or more derivatives with respect to a particular variable, the variable is called an independent variable. A variable is called dependent if a derivative of that variable occurs. In

OBJECTIVES: After completing this topic, you will be able to:

1. define differential equation.2. differentiate dependent to independent variables.3. identify the order and degree of differential equations.4. define a solution.5. discuss the types of solutions.6. identify a linear and nonlinear differential equations.

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the equation

Ld2 idt 2 +R

didt

+ 1Ci=Eωcost +

i is the dependent variable, t is the independent variable and L, R, C, E and are called parameters.The equation

d2Vdx2 + d2V

dy2 =0

has one dependent variable V and independent variable x and y.3. Ordinary differential equations are those in which all the differential coefficients

have reference to a single independent variable.Examples:

(1) dydx

=1−x2− y

(2) ( d3 ydx3 )

2

−dydx ( d2 y

dx2 )5

=f (x , y) =

(3) d2 y

dx2 −2 [1+( dydx )2]

4 /3

=0

4. Partial differential equations are those wherein there are two or more independent variables and partial differential coefficients with reference to any of them.

Example:

(4) x∂u∂ x

+ y∂u∂ y

=0

(5) ( ∂u∂ t )2

−k ( ∂2u∂ x2 + ∂2u

∂ y2 )5

=f (x , y )

5. The “order” of a D.E. is the same as the order of the highest-order derivative in

the equation.Example:

Equation (1) first order(2) third order(3) second order(4) first order(5) second order

6. The “degree” of a D.E. is the same as the power (or exponent) to which the derivative of the highest order in the equation raised.

Example:

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Equation(1) first degree(2) second degree(3) third degree(4) first degree

(5) first degree7. The solution to a D.E. is the relation involving only the variables and arbitrary

constants free from any derivatives or differentials, and consistent with, or satisfying, the given D.E.

8. Kinds of Solutions: There are three kinds of solutions of differential equations. They are:

1. The general solution, which is the solution containing a number of arbitrary constants equal to the order of the equation.

2. The particular solution, which is the solution obtained from the general solution by giving particular values to the constants.

3. The singular solution, which is the solution not containing any arbitrary constant and is not deducible from the general solution by giving particular values to the arbitrary constants in it, save in exceptional cases.

Arbitrary constants maybe obtained from known “boundaryconditions.”

9. The functional notation for the nth-order D.E. of the ordinary type is:

F ( dn ydxn ,

dn−1 ydxn−1 ,……

dydx

, x , y )=0

10. A “first-ordered D.E.” is written in either of the following common forms:

1. Implicit-Derivative: F ( dydx , x , y )=0

2. Implicit-Differential: M (x , y ) dx+N ( x , y )dy=0

3. Explicit-Derivative: dydx

=F(x , y )

11. Considering the differential equation as derived from the general solution, the latter is called the complete primitive of the former.

12. An equation which is linear, i.e., of the first degree in the dependent variable and its derivatives, is called linear differential equation. From this definition it follows that the most general (ordinary) linear differential equation of order n is of the form:

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Eq. 1 p0 (x)y(n) + p1(x) y( n - 1) + . . . + pn – 1(x) y’ + pn (x) y = r (x)

A differential equation which is not linear, i.e., cannot be put in the form of Eq. 1, is said to be nonlinear. In general, linear equations are much easier to solve than nonlinear ones, and most elementary applications involve linear equations.

Examples:1. The equation y” + 4xy’ + 2y = cos x is a linear

equation of the second order. The presence of the terms x, y’ and cos x does not alter the fact that the equation is linear because, by definition, linearity is determined solely by the way the dependent variable y and its derivatives occur in the equation.

2. The equation y” + 4yy’ + 2y = cosx is a nonlinearequation because of the occurrence of the product y and

one of its derivatives.

3. The equation y” + sin y = 0 is nonlinear because of the presence of sin y, which is a nonlinear function of y.

13. Initial Value Problem: By an initial value problem for an nth-order differential equation

F (x , y , dydx ,…. ,dn ydxn )=0

we mean: Find a solution to the differential equation on an interval I that satisfies at x0 the n initial conditions

y(x0) = y0

dydx

(x0 )= y1

.

.dn−1 ydxn−1 ( x0)= yn−1

where x0 I and y0, y1, . . . . , yn-1 are given conditions.

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In the case of a first-order equation, the initial conditions reduce to the single requirement

y(x0) = y0 ,

and in the case of a second-order equation, the initial conditions have the form

y(x0) = y0 , dydx

(x0 )= y1

The terminology initial conditions comes from mechanics, where the independent variable x represents time and customarily symbolized as t. then if t0 is the starting time, y(t0) = y0 represents the initial location of an object and y”(t0) gives its initial velocity.

Examples:

1. Show that (x) = sin x – cos x is a solution to the initial value problem

d2 ydx

+ y=0 ; y (0 )=−1 ,dydx

(0 )=1

Solution: Observe that (x) = sin x – cos x , dydx

=cos x+sin x

d2

dx2=¿ - sin x + cos x are all defined on (-, ). Substituting into the

differential equation gives

(-sin x + cos x) + (sin x – cos x) = 0

which holds for all x (-, ). Hence, (x) is a solution to the given differential equation on (-, ). When we check the initial conditions, we find

(0) = sin x – cos 0 = -1

ddx

(0 )=cos0+sin 0=1

which meets the requirements of the given equation. Therefore, (x) is a solution to the given initial value problem.

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Exercise No. 2

Determine the dependent and independent variables of the following:

1. x2y” + xy’ + (x2 – 4) y = 0

2.∂4u∂ x4 + ∂4u

∂ x2∂ y2 + ∂4u∂ y4 =0

3.∂2u∂ x2 + ∂2u

∂ y2 + ∂2u∂ z2 =∅ (x , y , z)

4. xd2 ydx2 − y ( dydx )

2

=4

5. (y”)2 + 4y’ + 3y = x

6. (x + 2) dy = (1 – 2x) dx

7. sinx( d2 ydx2 )

3

−e x( dydx )5

=xy

8. xd4 ydx4 +5 y

d2 ydx2 −( dydx )

6

=0

9.∂2 z∂ t2 −3

∂2 y∂ t2 −∂2 z

∂u2 =0

10. ( ∂2 y∂ t 2 )

2

−7( ∂2 y∂t 2 )

4

=t+u

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.

.

.Exercise No. 3

Describe each of the following giving its order and degree, and telling whether it is ordinary or partial and linear or nonlinear.

Order Degree Ordinary/Partial Linear/Nonlinear2. y” + 3y’ + 2y = x4

3. y’’’ + 6y” + 11y’ + 6y = ex

4.d (xy ' )dx

+xy=0

5. a2 ∂2u

∂ x2 =∂2u∂ t 2

6.∂2u∂ x2 =u

∂u∂ t

7. (y”)2 + 4(yy’)2 – xy = 0

8. x4(y’)4 + (x – 1) y’’’ = sin x

9. ( d2 sdt 2 )

4

−dsdt

=sint+cost

10.drdθ

+r=2θ

11.dn ydxn + y3−1=0

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Unit 2

FORMATION OF DIFFERENTIAL EQUATIONS

2.1 An inverse Problem: Elimination of Arbitrary Constants

2.1.1 By Differentiation and Combination

One of the ways differential equations are obtained is the elimination of arbitrary constants from a given equation. The general method to eliminate n arbitrary constants is to differentiate the equation n times.

The n arbitrary constants may then be eliminated from the n + 1 equations. The result will be differential equations of order n.

Eliminate the arbitrary constants from the given equations by combining after differentiation.

Example 2-1 y = c1 cos 3x + c2 sin 3x.

Solution: y’ = -3c1 sin 3x + 3c2 cos 3xy” = -9c1 cos 3x - 9c2 sin 3x = -9(c1 cos 3x + c2 sin 3x)

= - 9yy” + 9y = 0 Ans.


obtain the differential equation by eliminating constants using differentiation and elimination, isolation of constants, and elimination by determinants.

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Example 2-2 y = x + c1ex + c2e-2x (1)Solution: y’ = 1 + c1ex - 2c2 e-2x (2)

y” = c1 ex + 4c2 e-2x (3)

Subtract Eq. (1) from Eq. (2).y’ - y = 1 - x -3c2 e-2x (4)

Subtract Eq. (2) from Eq. (3).y” - y’ = -1 + 6c2 e-2x (5)

Multiply Eq. (4) by 2.2y’ - 2y = 2 - 2x -6c2 e-2x (6)

Adding Eqs. (6) and (5), we obtain

y” + y’ - 2y = 1 - 2x Ans.

Example 2-3 y = c1e-x + c2e2x + c3e-3x (1)

Solution: y’ = - c1e-x + 2c2e2x - 3 c3e-3x (2)Adding Eqs. (1) and (2),

y’ + y = 3c2e2x - 2c3e-3x

y” + y’ = 6c2e2x + 6c3e-3x

= 6 (c2e2x + c3e-3x) = 6y - 6c1 e-x (3)

y’’’ + y” = 6y’ + 6c1e-x (4)Adding Eqs. (3) and (4), we have

y’’’ + 2y” + y’ = 6y’ + 6y

y’’’ + 2y” - 5y’ - 6y = 0. Ans.

Example 2-4 y = ex(c1cos 2x + c2sin 2x) (1)

Solution: e-x y = c1cos 2x + c2sin 2x (2)

Differentiate Eq. (2).e-x y’ - e-xy = -2c1 sin 2x + 2c2 cos 2x

Differentiating again, we get e-x y” - e-xy’ - e-x y’ + e-xy

= - 4( c1cos 2x + c2sin 2x)= - 4e-x y

y” - 2y’ + y = - 4y

y” - 2y’ + 5y = 0 Ans.

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Example 2-5 y = c1 sin at + c2 cos at,(a not to be eliminated)

Solution: y = c1 sin at + c2 cos aty’ = a c1 cos at - a c2 sin aty” = - a2 (c1 sin at + c2 cos at) = - a2 yy” + a2 y = 0. Ans.

Example 2-6 y = Ax2 + Bex (1)

Solution: Differentiating twice, we obtainy’ = 2Ax + Bex (2)y” = 2A + Bex (3)

Subtracting Eq. (1) from Eq. (2),y’ - y = 2Ax - Ax2 = A(2x - x2) (4)y” - y’ = 2A - 2 Ax = A(2 – 2x) (5)

Dividing Eq. (5) by Eq. (4), we have

y -y'} over {{y} ^ {'} - y} = {A(2-2x)} over {A(2x- {x} ^ {2} )¿

(2x - x2)y” + (x2 - 2)y’ + 2(1-x) y = 0 Ans.

2.1.2 By Isolation of Constants

In this method at each step before differentiation, isolate one of the arbitrary constants, free from any variable, so that it will disappear after differentiation.

For the following problems, eliminate the arbitrary constants by isolation.

Example 2-7 x sin y - xy2 = c

Solution: Differentiation once will eliminate the constant c.

x cos y y’ + sin y - 2xy y’ - y2 = 0(x cos y - 2xy) y’ - y2 + sin y = 0 Ans.

Example 2–8 y = ax2 + bx + c

Solution: y’ = 2ax + by” = 2ay’’’ = 0 Ans.

Example 2-9 y = Ax2 + Bex

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Solution: Isolate B by multiplying the entire equation by e-x.

e-x y = Ax2 e-x + Be-x y’ - e-x y = - Ax2 e-x + 2Ax2 e-x y’ - y = - Ax2 + 2Ax = A (-x2 + 2x)

A = y '− y

−x2+2 x

Differentiating again, we get

0 = (−x 2+2x )( y ”– y’)−( y ’ – y)(−2x+2)

(−x 2+2x )2

(2x - x2) y” + (x2 - 2) y’ + 2 (1 – x) y = 0 Ans. Example 2-10 y = x + c1ex + c2e-2x

Solution: e2x y = xe2x + c1e3x + c2

e2x y’ + 2e2x y = 2xe2x + e2x + 3c1e3x y’ + 2y = 2x + 1 + 3c1 ex

e-x y’ + 2 e-x y = 2x e-x + e-x + 3c1 e-x y” - e-x y’ + 2 e-x y’ - 2 e-x y

= - 2 xe-x + 2e-x - e-x

y” + y’ - 2y = 1 - 2x. Ans.

Example 2-11 x2 y = 4 + cy

Solution: c=x2− 4y

; 0=2x−(−4 y '

y2 )4y’ + 2xy2 = 02y’ + xy2 = 0 Ans.

2.1.3 Elimination By Determinants

A third method of eliminating arbitrary constants is based on a theorem in Algebra:In order for a system of n + 1 linear equations in n unknowns to be consistent, the determinant formed from the coefficients of the unknowns and the terms free of the unknowns must vanish.

The determinant formed is called the eliminant of the arbitrary constants.

For the following problems, eliminate the arbitrary constants using determinants.

Example 2-12 y = Ax2 + Bex

Solution: Differentiate the equation twice.y’ = 2Ax + Bex

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y” = 2A + Bex

Form the determinant whose elements are the coefficients of the arbitrary constants A and B, and the terms y, y” and y’’’, which are free of the arbitrary constants.

[ y x2 e x

y ' 2 x e x

y # 2 # {e} ^ {x}} right ] =0 or left [matrix {y # {x} ^ {2} # 1 ## y' # 2x # 1 ## y 2 1 ]=0

Since ex is a common factor of the term in the third column. Expanding the determinant, we obtain

(2x - x2)y” + (x2 - 2) y’ + (2 - 2x) y = 0. Ans

Example 2-13 y = c1 cos 3x + c2 sin 3x

Solution: Differentiate the equation twice.y’ = -3c1 sin 3x + 3c2 cos 3xy” = - 9c1 cos 3x - 9c2 sin 3x

Form the determinant

¿

Expanding the determinant and simplifying we obtain

y” + 9y = 0 Ans.

Example 2-14 y = x + c1ex + c2e-2x

Solution: Differentiate the equation twice.y’ = 1 + c1ex - 2c2e-2x

y’ = c1ex + 4c2e-2x

Write the equations asy = x + c1ex + c2e-2x

y’ = 1 + c1ex - 2c2e-2x

y’ = c1ex + 4c2e-2x

Form the determinant

¿

22

Expanding the determinant and simplifying, we get

y” + y’ - 2y = 1 - 2x. Ans.

Example 2-15 y = c1 x2 + c2x3

Solution: Differentiate the equation twice.

y’ = 2c1 x + 3c2x2

y” = 2c1 + 6c2xThe determinant in this case will be

¿

Expanding the determinant the result is

x2 y” - 4xy’ + 6y = 0. Ans.

Exercise No. 4

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Determine the differential equations of the following by eliminating constants using any method.

1. x sin y – y cos x = c

2. y = c1 + c2e2x

3. y = c1x + c2

x

4. y = c1x2 + c2x3

5. y = c1cos 2x + c2 sin 2x

6. y = c1 x + c2 ex

7. y = c1 e2x + c2 x e2x

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8. y = c1 ex + c2 x e2x + c3 x2 ex

9. y = e2x(c1 cos x + c2 sin x)

10. y = c1 ex + c2e-x + c3 x2 e-x

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2.2 Differential Equation of A Family of Curves

A family of curves of curves on a plane is usually defined by an equation containing one or more parameters together with the coordinates of a point on the plane. The differential equation of the family is obtained by eliminating

the parameters by the methods in Art. 2-1

In the following problems obtain the differential equations of the given families of curves whose properties are indicated.

Example 2-16 All straight lines through the fixed point (h, k)

Solution: The equation of a line through the point (h, k) is

y - k = m(x - h)y’ = m.

Hence, the required differential equation is

y - k = (x - h) y’ Ans.

Example 2-17 All straight lines

Solutions: The equation of a line whose slope is m and whose y-intercept is b is

y = mx + bHere m and b are arbitrary constants. Differentiating twice, we have

y’ = m ; y” = 0 Ans.

Example 2-18 Straight lines with equal x- and y-intercepts.


determine the differential equation of the given family of curves from indicated properties.

a

b

a (0, k)

26

Solution: The figure shows a straight line with x- and y-intercepts a and b, respectively. Using the intercept form of the equation of a straight line, we have

xa

+ yb

=1. But a = b

Hence, x + y = a. 1 + dydx

= 0

dx + dy = 0 Ans.

Example 2-19 Circles with center on the y-axis.

Solution: Let the center be at (0, k) and radius = a

x2 + (y – k) = a2

2x + 2(y – k) y’ = 0

y - k = −xy '

y '=− y '−xy } over {{y'} ^ {2} ¿

y’3 = -y’ + xy”

xy” - y’3 - y’ = 0 Ans.

Example 2-20 Circles tangent to the y-axis.

Solution: Let (h, k) be the coordinates of the center of the circle. Since the circle is tangent to the y-axis its radius r = h. The general equation is

(x - h)2 + (y - k)2 = h2 (1)x2 - 2hx + (y - k)2 = 0

. (h, k)

F

(0, a)

(0, -a)

F

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2h = x + ( y−k )2

x (2)

Differentiating Eq. (2), we have

0 = 1 + x ∙2 ( y−k ) y '−( y−k )2

x2

(y – k)2 - 2xy’ (y – k) - x2 = 0

y – k = xy’ x√ y ' 2+1 (3)

Differentiating Eq. (3), we obtain

y’ = xy” + y’ x ∙2 y ' y } over {2 sqrt {{y'} ^ {2} + 1}¿ √ y ' 2+1

after some simplification, we obtain

xy” √ y ' 2+1 xy’y” (y’2 + 1) = 0. Ans.

Example 2-21 Parabolas with the vertex and focus on the y-axis.

Solution: Let the vertex be at (0, k). The general equation isx2 = 4a( y k)2x = 4ay’ ; x = 2ay’ ; 1 = 2ay”x1

= 2ay '2ay} = {y'} over {y

xy” - y’ = 0 Ans.

28

Exercise No. 5

Name: ____________________________ Score: ___________ Rating: ___________

In the following problems obtain the differential equation of the given family of curves whose properties are indicated.

1. All circles passing through the origin and (0, 4).

2. Straight lines with slope and y-intercept equal.

3. Straight lines with algebraic sum of the intercepts fixed as k.

29

4. All circles tangent to both x and y axes.

5. The cissoids y2 = x3

c−x

6. The strophoid y2 = x2(a+x)a−x

7. The hyperbolas b2x2 - b2y2 = a2b2.

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8. Parabolas with vertex and focus on the x-axis.

9. The folium of Descartes x3 + y3 = 3axy.

10. The trisectrices of Maclaurin y2 (a + x) = x2

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Unit 3

SOLUTIONS TO FIRST-ORDERED DIFFERENTIAL EQUATIONS

3.1 Separation of Variables

1. If the variables in M(x, y) dx + N (x, y) dy = 0 can be separated into the form

f(x) dx + g(y) dy = 0the solution is by integration

∫ f(x) dx + ∫ g(y) dy = c

2. TEST: The variables are separable if the functions M and N are factorable as follows:

M(x, y) = F(x) G(y)N(x, y) = H(x) (y)

Find the complete the solutions of the following D.E.:


1. classify the different forms of first-ordered differential equations: separation of variables, homogeneous D.E., exact D.E., non-exact D.E., first-ordered linear D.E., Bernoulli equation, D.E. linear in x and y.

2. identify a variable-separable differential equation.3. distinguish a homogeneous differential equation.4. test the exactness of a differential equation.5. write the given D.E. in the standard form of Bernoulli equation and

first-ordered linear differential equations.

32

Example 3-1 xcos2 y dx + tan y dy = 0

Solution: xcos2 y dx + tan y dy = 0 is reduced by dividing each termby cos2 y

x dx + tan y sec2 y dy = 0

Both terms are integrable by the “power” formula,

x2

2+ tan2 y

2=c or x2 + tan2 y = c1 Ans.

Example 3-2 xy3ln x dx - 2 dy = 0

Solution: Division by y3 separates the variables into,

x lnx dx - 2 y-3 dy = =

The second term is integrated by the “power” formula, while the first is by “integration by parts”,

Let u = lnx du = dx/x dx = x dx v = x2/2

Substitution gives,x2

2lnx−∫ x2

2dxx

−2∫ y−3dy=c

andx2

2lnx− x2

4+ y−2=c

Simplification gives,

X2(ln x2 - 1) + 4y-2 = c1 Ans.

Example 3-3 y’ = 2x+2 x y2

y+2x2 y

Solution: The given D.E. may be written as,

dydx

=2 x+2x y2

y+2x2 y

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and the variables are separated into,

2 xdx

1+2x2− ydy

1+ y2=0

Integration leads to the ln function

12

ln (1+2 x2)− 12

(1+ y2 )=c

Further simplification gives,

1+2x2

1+ y2 =c1 Ans.

34

Exercise No. 6

(Separation of Variables)

1. dy = tan x tan y dx

2. ex+y dx + (1 + ex) dy = 0

3. ( 4+9 x2 )dy−√9−4 y2 dx = 0

35

4. (xy – x) dx = (x2y2 - x2 + y2 - 1) dy

5. (xy + x) dx + (1 + x2)dy = 0

6. 2ex cot y dx - (4 + ex) csc2 y dy = 0

36

7. dydx

=(x+1 )( y−2)

xy

8. xlnx lny dy + dx = 0

9. 2ex cot y dx + (1 + ex) csc2 y dy = 0

10.dydx

= sin(x + y)

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3.2 Homogeneous Differential Equations

(a) In M(x, y) dx + N(x, y) dy = 0, if

M(rx, ry) = rn M(x, y) and

N(rx, ry) = rn N(x, y)

Such that n = m, then the D.E is said to be “homogeneous” of “degree n”. Here, r can either be a variable or a constant.

(b) The solution of such type of D.E is by either of the following substitution methods:First: Let y = vx

dy = v dx + x dv

Here, y is eliminated and v introduced into the D.E. In the final analysis, the D.E. takes the forms in which h the variables are separable. After the solution, the variable y is substituted back to the equation.

Second: Let x = uy dx = u dy + y du

Solution is similar to that of the previous method, except that here x is eliminated.

(c) Suggested Solution

If M is simpler than N: Use x = uyIf N is simpler than M: Use y = vx

For the following D.E., find the solution:

Example 3-4 (x3 + y3) dx - 3xy2 dy = 0

38

Solution: (x3 + y3) dx - 3xy2 dy = 0 is “homogeneous of degree 3”

Here N = -3xy2 is simpler than M = x3 + y,

Use: y = vx and dy = vdx + x dv

Substitute to the given D.E.,

(x3 + v3x3) dx - 3x (v2 x2) (vdx + x dv) = 0

Divide each term by x3 and simplify(1 + v3) dx - 3v2 (vdx + x dv) = 0and

d xx

− 3 v2dv1−2v3 =0

Integration leads to the ln functions,

ln x + 12

ln (1−2v3 )=c

Since y = vx and v = y/x , then

ln x + 12

ln(1− 2 y3

x3 )=c

Simplification finally gives,

x3 - 2y3 = c1x Ans.

Example 3-5 xy dx - (x + 2y)2dy = 0

Solution: xy dx - (x + 2y)2dy = 0 is “homogeneous of degree 2”.

Here, M xy is ismpler than N = -(x + 2y)2

Use: x = uy and dx = udy + ydu

Substitute to the given D.E.,

uy2 (udy + ydu) - (uy + 2y)2dy = 0

Divide each term by y2 and simplify,

39

u2 dy + uy du - (u + 2)2dy = 0u2 dy + uy du - (u2 + 4u + 4)dy = 0

and uy du - 4(u + 1) dy = 0

Separation of variables gives,udu

4 (u+1)−dy

y=0

14 (1− 1

u+1 )−dyy

=0

Integrate: 14

[u−ln (u+1 ) ]−lny=c

Substitute x = uy or u = x/y and simplify,

x - y ln (xy3 + y4) = c1y Ans.

Example 3-6 (x + y sin y/x) dx - x sin y/x dy = 0

Solution: (x + y sin y/x) dx - x sin y/x dy = 0 is “homogeneous of degree 1”.

Use: y = vx or v = y/x

dy = vdx + x dv

Substitute to the D.E.,(x + vx sin v) dx - x sin v (vdx + x dv)= 0

Divide each term by x and simplify,dx - x sin v dv = 0

and dxx

−sinvdv=0

Integration gives,ln x + cos v = c

Finally, using v = y/x,

40

ln x + cos y/x = c Ans.

Exercise No. 7

(Homogeneous Differential Equations)

Find the general solution of the following differential equations.

1. (4x2 - y2) dx - xy dy = 0

2.dxdy

= x2

y2 +2x

41

3. (3xy – 2y2) dx + (2x2 - 3xy) dy = 0

4. ( dydx− yx ) ln

yx

= xy

42

5. (sinyx ) dydx= x

ysin

yx

+1

6. x dy - y dx = √ x2+ y2 dx

7. (2x - y) dx - (x + 4y) dy = 0

43

8. (x + y sin xy

) dx - (y + x sin xy

) dy = 0

9. Find the particular solution of (y2 + 2xy) dx + 2x2dy = 0; x = 1; y = -2

10. Find the particular solution of y(x – 6y) dx - x (2x – 9y) dy = 0 if y(0) = 1

44

3.3 Exact Differential Equation

(a) The equation M(x, y) dx + N(x, y) dy = 0 is said to be “exact” if

∂M∂ y

=∂ N∂ X

(b) Solutions:

(i) Direct: ∫Mdx+∫(N− ∂∂ y

∫Mdx)dy=0

(ii) Indirect: ∫ Mdx + f(y) = cand ∫ Ndy + g(x) = c

REMARKS: (a) The unknown functions f(y) and g(x) are obtained from the comparison of these two (2) possible solutions.(b) In integrating the functions M and N, “ partial”

integration is applied, that is, to hold y constant when integrating relative to x and to hold x constant in integrating with respect to y. Same is true in partial differentiation.

Find the solutions of the following D.E.

Example 3-7 (2xy - 3x2)dx + (x2 + 2y) dy = 0

45

Solution: (2xy - 3x2)dx + (x2 + 2y) dy = 0 is first tested for “exactness”.

M = 2xy – 3x2 ; ∂M∂ y

=2 x

N = x2 + 2y ; ∂N∂ X

=2x

So, the given D.E. is exact.

Using the Direct Method,

∫Mdx+∫(N− ∂∂ y

∫Mdx)dy=0

∫(2 xy−3 x2)dx+∫ [ x2+2 y− ∂∂ y

∫(2xy−3 x2)dx ]dy=0

x2 y−x3+∫ [ x2+2 y− ∂∂ y

∫( x2 y−x3)] dy=0

x2 y – x3 + ∫ (x2 + 2y - x2) dy = c

Finally, x2 y – x3 + y2 = c Ans.

Example 3-8 (cos y + y cos x) dx - (x sin y - sin x) dy = 0

Solution: In the given D.E.,

M = cos y + y cos x∂M∂ y

= - sin y + cos x and

N = -x sin y + sin x∂N∂ X

= - sin y + cos x

Therefore, the D.E. is exact,

Using the Indirect Method,

(1) ∫ M dx + f(y) = c ∫ (cos y + y cos x) dx + f(y) = c

46

x cos y + y sin x + f(y) = c

(2) ∫ N dy + g(x) = c ∫ (-x sin y + sin x) dy + g(x) = c x cos y + y sin x + g(x) = c

Comparison of the two (2) possible solutions (1) and (2) shows that f(y) = g(x) = 0, so

x cos y + y sin x = c Ans.

Example 3-9 (ex + ln y + y/x) dx + (x/y + ln y + sin y) dy = 0

Solution: Here, the D.E. has

M = ex + ln y + yx

: ∂N∂ X

= 1y

+ 1x

N = xy

+ ln x + sin y ; ∂N∂ X

= 1y

+ 1x

Use the Indirect Method

(1) ∫ M dx + f(y) = c

∫ (ex + ln y + yx

)dx + f(y) = c

ex + xlny + ylnx + f(y) = c

(2) ∫ N dy + g(x) = c

∫( xy

+ ln x + sin y) dy + g(x) = c

x ln y + y lnx - cos y + g(x) = c

Comparison of solutions (1) and (2) shows thatf(y) = - cos y and g(x) = ex, so

ex + xlny + ylnx - cos y = c

47

Exercise No. 8

Exact Differential Equations


1. (2xy - 6x2) dx + (x2 - y) dy = 0

2. (2x + 3y) dx + (3x - y + 5) dy = 0

48

3. 3y (x2 - 1) dx + (x3 + 8y - 3x) dy = 0

4. (3x2 - 4y2) dx - (3y2 + 8xy) dy = 0

5. (2xy – 3y) dx + (x2 - 3x) dy = 0

6. (3x2y - sin y) dx + x(x2 – cos y) dy = 0

49

7. ( 1- 2xy + y2) dx + (2xy – x2) dy = 0

8. (x - 2y) dx + 2(y - x) dy = 0

9. (y2 - 2xy + 6x) dx - (x2 - 2xy + 2) dy = 0

50

10. (6x + y2) dx + y(2x – 3y) dy = 0

3.4 Non-Exact Differential Equations

(a) The equation M(x, y) dx + N(x, y) dy = 0

∂M∂ y

≠∂ N∂ X

then the D.E. is “non-exact”.

(b) Solution: A “non-exact” D.E. is solved by other methods or more often by reducing it to an “exact” form through the determination of an “integrating factor” which, when multiplied to the D.E., reduces it to

M Φ dx + N Φ dy = 0 such that

∂MΦ∂ y

=∂ N Φ∂ X

where: Φ = Φ(x, y) is the integrating factor.

(c) Special Cases for Finding Φ

CASE 1: When Φ is a function of x alone,

51

Φ = Φ(x)

TEST: If 1N ( ∂M

∂ y−∂N

∂ x )=f (x)

Then Φ = e∫ f(x) dx.

CASE 2: When Φ is a function of y alone,

Φ = Φ(x)

TEST: If 1M ( ∂ N

∂ x−∂M

∂ y )=g( y )

Then Φ = e∫ g(y) dy.

CASE 3: When Φ = xm yn

TEST: If Mny

−Nmx

=∂N∂ x

−∂M∂ y

reduces to an “identity” equation (one wherein the terms at the left side are the same as on the right side) the exponents m and n may be obtained by the comparison of terms from both sides of the equation - that is, by equating coefficients of like terms.


Example 3-10 (x2 + y2 + 1) dx + (x2 - 2xy) dy = 0

Solution: M = x2 + y2 + 1 ; ∂M∂ y

=2 y

N = x2 - 2xy ; ∂N∂ x

= 2x - 2y

so, D.E. is non-exact.

This is under CASE 1 with the “integrating factor”,

Φ = e∫ f(x) dx

52

f ( x )= 1N ( ∂M

∂ y−∂ N

∂ x )=

1

x2−2xy (2y - 2x + 2y)

= 2(−x+2 y)x (x−2 y ) or

−2x

So, Φ = e∫ (-2/x) dx = e-2lnx = x-2

Multiply the given D.E. by Φ = x-2

(1 + x-2 y2 + x-2) dx + (1 - 2x-1 y) dy = 0The new functions M and N are,

Mn = 1 + x-2 y2 + x-2 ; ∂M n

∂ x = 2 x-2 y

Nn = 1 - 2x-1 y ; ∂N n

∂ y = 2x-2y

So, the reduced D.E. is now exact

Use the Indirect Method

(1) ∫ Mn dx + f(y) = c ∫ (1 + x-2 y2 + x-2 ) dx + f(y) = c

x - x-1 y2 - x-1 + f(y) = c

(2) ∫ Nn dy + g(x) = c ∫( 1 - 2x-1 y ) dy + g(x) = c

y - x-1y2 + g(x) = C

Comparison of solutions (1) and (2) shows thatf(y) = y and g(x) = x - x-1, so

x - x-1y2 - x-1 + y = c Ans.

Example 3-11 (2xy - y2 + y)dx + (3x2 - 4xy + 3x ) dy = 0

Solution:

53

M = 2xy - y2 + y; ∂M∂ y

= 2x - 2y + 1

N = 3x2 - 4xy + 3x ; ∂N∂ x

= 6x - 4y + 3

so, D.E. is non-exact.

This D.E. can be solved under CASE 2 with the “integrating factor”.Φ = e∫ g(y) dy

where: g ( y )= 1M ( ∂ N

∂ x−∂M

∂ y )

¿1

2xy− y2+ y(4 x−2 y+2 )

¿2(2x− y+1)y (2x− y+1) =

2y

So, Φ = e∫ (2/y) dy = e2lny = y2

Multiply each term in the given D.E. by Φ = y2

(2xy3 - y4 + y3)dx + (3x2 y2 - 4xy3 + 3x y2 ) dy = 0

The new M and N functions are,

Mn = 2xy3 - y4 + y3 ; ∂M n

∂ y = 6xy2 - 4y3 + 3y2

Nn = 3x2 y2 - 4xy3 + 3x y2 ; ∂N n

∂x = 6xy2 - 4y3 + 3y2

Having been reduced to an exact form , the D.E, may now be solved either by the Direct or Indirect Method, to give the final answer,

x2y3 - xy4 + xy3 = c Ans.

Example 3-12 2y dx + (x - x3y3) dy = 0

Solution: M = 2y ; ∂M∂ y

=2

54

N = x - x3y3 ; ∂N∂ x

= 1 - 3x2y3

This non-exact D.E. falls under Case 3 where the “integrating factor” is

Φ = xm yn

with m and n obtained from

Mny

−Nmx

=∂N∂ x

−∂M∂ y

2 yny

–(x−x3 y3 )m

x=1−3x2 y3−2

` 2n - m + mx2y3 = -1 - 3x2y3

Note that this is an “identity “ equation,

where : 2n - m = -1and m = -3so, n = -2 and Φ = x-3y-2

Multiply the given D.E. by Φ = x-3y-2,

2x-3 y-1 dx + (x-2y-2 - y) dy = 0

The new M and N functions are,

Mn = 2x-3 y-1 ; ∂M n

∂ y = - 2x-3 y-2

Nn = x-2y-2 - y ; ∂N n

∂x = - 2x-3y-2

This is now an exact D.E. which can be solved by either the Direct or Indirect method, to give

2 x-2y-1 + y2 = c Ans.

55

Exercise No. 9

Non-Exact Differential Equations


1. y(4x + y) dx - 2(x2 - y) dy = 0

56

2. 3(x2 + y2) dx + x(x2 + 3y2 + 6y) dy = 0

3. 2(x – y + 2) dx + x(x – 2y + 2) dy = 0

4. y(2x + y – 2)dx – 2 (x + y) dy = 0

5. (xy + 1) dx + x(x + 4y – 2)dy = 0

57

6. 2y (x2 – y + x) dx + (x2 – 2y) dy = 0

7. (4x + y) dx – 2(x2 – y) dy = 0

8. (2y2 + 3xy – 2y + 6x) dx + x(x + 2y – 1) dy = 0

58

9. (2y2 + 5xy – 2y + 4) dx + x(2x + 2y –1) dy = 0

10. (x2 – y2 – 4y – 1) dx + 2x(x + y + 2) dy = 0

3.5 First-Ordered Linear Differential Equations (FOLDE)

(a) This is a D.E. which is “first order” in the derivative and “linear” in the dependent variables. In the standard form, it is written as

dydx

+ y P(x) = Q(x)

dx(b) Solution: y Φ = ∫ Q(x) Φ dx + c

where: Φ = e∫ P(x) dx

NOTE: FOLDE is actually a “non-exact” D.E. where Φ is obtained under CASE 1.

REMARK: For the case where the dependent variable is x, FOLDE is

59

written in its standard form as

dxdy

+ x P(y) = Q(y)

having a solution: x Φ = ∫ Q(y) Φ dy + c

where: Φ = e∫ P(y) dy


Example 3-13 dydx

+ 2xy - 4x = 0

Solution: The standard form of this D.E. in the FOLDE form is,

dydx

+ 2xy = 4x

where: P = 2x ; Q = 4x

Φ = e∫ Pdx = e∫ 2x dx = ex2

Solution is: y Φ = ∫ Q Φ dx + c

yex2

=∫ ( 4 x ) ex2

dx+c

Integration gives,yex2

=2ex2

+c

y=2+ce− x2

Ans.

Example 3-14 t ds - (t3 e3t + 3st + s) dt = 0

Solution: Divide the D.E. by tdt and reduce to the FOLDE form,

dsdt

- s (3 + t-1) = t2 e3t

where: P = - (3 + t-1)Q = t2 e3t

and Φ = e∫ Pdt = e−(3+t ¿¿−1)dt ¿

= e-3t - ln t or t-1 e-3t

The solution is,

60

s Φ = ∫ Q Φ dt + cs t-1 e-3t = ∫ t2 e3t (t-1 e-3t) dt + c

Integration leads to,

s t-1 e-3t = t2

2 + c

and s = 12

t3 e3t + c t e3t. Ans.

Example 3-15 y dxdy

- 2x - 3y2 + 2 = 0

Solution: Divide the given D.E. by y,dxdy

- 2y

x = 3y - 2 y-1

which is now in the FOLDE form with,

P = - 2y

; Q = 3y - 2y-1

and Φ = e∫ P dy = e∫ -(2/y) dy = e-2 lny

Φ = y-2

Solution is: x Φ = ∫ Q Φ dy + c

x y-2 = ∫ (3y - 2y-1) y-2 dy + cIntegration gives,

x y-2 = 3 ln y + y-2 + c

which simplifies to,x = y2 ln y3 + 1 + cy2 Ans.

61

Exercise No. 10

First-Ordered Linear Differential Equations



1.dydx

+ cot x = 3ecos x

62

2.dydx

+ ( 1x

+ tan x ) y = 2 sec x

3. (1 - x2)dydx

- xy = 3x (1 - x 2)

4. ydx = (ey + 2xy - 2x) dy

63

5.dydx

- yx

= x+ yx−1

6.dydx

= 2x – 2y cot 2x

7. x dydx

- (x + 2)2 = y

8.dydx

- 1

xlnx y =

1

x2 lnx

64

9. xdy + (x2 – 3y) dx = 0

10. y’ + (2x + 1) y = e− x2

3.6 The Bernoulli Equation

(a) The Bernoulli equation is a “non-linear” D.E. having the standard form

dydx

+ y P(x) = yn Q(x)

(b) The Bernoulli equation is reducible to a “linear” form (FOLDE) by making use of the substitution

65

z = y1 – n

which givesdydx

= yn

1−n dzdx

With this, the Bernoulli equation is reduced to

dzdx

+ (1 - n) z P (x) = (1 - n) Q (x)

or dzdx

+ z Pr (x) = Qr (x)

where: Pr (x) = (1 – n) P(x)Qr(x) = (1 – n) Q(x)

(c) Solution: y1 – n Φ = ∫ Qr(x) Φ dx + c

where: Φ = e∫ Pr (x) dx

REMARK: For the case where the dependent variable is x, the Bernoulli equation is,

dxdy

+ x P(y) = xn Q(y)

Elements in the solution of this are similar to the previous discussion.

Example 3-16 y’ + yx

- 3x2y2 = 0

Solution: The Bernoulli form of the given D.E. is,

dydx

+ 1x

y = 3x2 y2

with n = 2 ; 1 - n = -1

66

P = 1x

; Q = 3x2

The solution is

y1 - n Φ = ∫ Qr Φ dx + c

where: Pr = (1 – n) P = - 1x

Qr = (1 – n) Q = -3x2

Φ = e∫ Pr dx = e∫ -(1/x) dx

= e-ln x = x-1

Substitution and integration gives,

y-1 x-1 = ∫ (-3x2) x-1 dx + c

y-1 x-1 = - 32

x2 + c Ans.

Example 3-17 3 dxdy

+ 3 xy

= 2x4y4

Solution: Divide the given D.E. by 3,

dxdy

+ 1y

x = 23

y4 x4

which is now in the Bernoulli form

P = 1y

; Q = 23

y4

n = 4 and 1 – n = - 3

The solution is

x1 - n Φ = ∫ Qr Φ dy + c

where: Pr = (1 – n) P = - 3

67

xQr = (1 – n) Q = - 2y4

Φ = e∫ Pr dy = e∫ -(3/y) dy

= e- 3 ln y = y –3

Substitution and integration leads to,

x-3y-3 = ∫ (-2y4) y-3 dy + cFinally,

x-3y-3 = - y2 + c Ans.

Example 3-18 2 wt dtdw

= t2 - 2w3

Solution: The given D.E. is divided by 2 wt,

dtdw

- 1

2w t = - w2 t-1

which is now in the Bernoulli form

Here, P = - 1

2w ; Q = - w2

n = - 1 and 1 – n = 2

The solution is

t1 - n Φ = ∫ Qr Φ dw+ c

where: Pr = (1 – n) P = - 1w

Qr = (1 – n) Q = - 2w2

Φ = e∫ Pr dw = e∫ -(1/2) dw

= e- ln w = w – 1

68

Substitution and integrate,

t2w-1 = ∫ (-2w2) w-1 dw + c

t2

w = - w2 + c Ans.

Exercise No. 11

69

Bernoulli Equation


1.dydx

- 2x

y = ( y3 )3

sin x

2.dydx

- y = 2 xy2 e-x

3. x dydx

= y + y3 cos x

70

4. dxdy

- x cot 2y = x3 csc 2y

5.dydx

- 2x

y = ( yx )3

cos x

6.dxdy

+ 4y

x = x4 y3

7. 3dydx

+ 3x

y = 2 x4 y4

71

8. 3xy’ + y + x2y4 = 0

9.dydx

+ yx

= √ y

10.dydx

+ y = 12e2x y

72

3.7 D.E. Linear in x and y

(a) The standard form for this 1st-ordered D.E. is

(A1x + B1 y + C1) dx + (A2x + B2 y + C2) dy = 0

where: C1 and/or C2 0

(b) Solutions:

CASE 1: When A1x + B1 y + C1 = 0 and A2x + B2 y + C2 = 0 represent two parallel lines,

that is, when:

A2 = k A1 and B2 = k B1

(k – constant of proportionality)]

USE: v = A1x + B1 y dv = A1 dx + B1 dy

and substitute to the D.E., eliminating either x or y.

NOTE: This substitution will reduce to one where the variables are separable.

CASE 2: When A1x + B1 y + C1 = 0 and

A2x + B2 y + C2 = 0 represent two non-parallel lines,

USE: x = u + h and dx = du y = v + k and dy = dv

where: (h, k) is the point of intersection of the two parallel lines.

NOTE: This substitution method will reduce the D.E. to one that is homogeneous of the first degree.

Example 3-19 (x + y - 1) dx + (2x + 2y + 1) dy = 0

Solution: This D.E. has the coefficients of dx and dy when equated to

73

zero (0) represent two (2) parallel

x + y - 1 = 0 and 2x + 2y + 1 = 0

under CASE 1. with k = 2

Let v = x + y or dv = dx + dy

and substitute to the D.E.,

(v – 1)(dv – dy) + (2v + 1) dy = 0

vdv – dv - vdy + dy + 2v dy + dy = 0

(v - 1)dv + (v + 2) dy = 0

v−1v+2

dv + dy = 0

Integrate after reducing the first term,

∫(1− 3v+2 )dv + y = c

v - 3 ln(v + 2) + y = c

Substitute v = x + y

x + 2y - ln (x + y + 2)3 = c

Example 3-20 (x - 4y - 9) dx + (4x + y - 2) dy = 0

Here, when the coefficients of dx and dy are equated to zero (0),

x - 4y - 9 = 0 and 4x + y - 2 = 0represent two (2) non-parallel lines with point of intersection at (1, -2). this falls under CASE 2.

Let x = u + 1 and dx = duy = v - 2 and dy = dv

and substitute to the given D.E.,

74

{ u + 1 - 4(v – 2) - 9} du + {4(u + 1) + v - 2 - 2} dv = 0

or (u - 4v) du + (4u + v) dv = 0

which is “homogeneous of degree 1”

Let u = zvdu = z dv + v dz

Substitute to get,

(zv – 4v) (z dv + v dz) + (4 zv + v) dv = 0

Divide by v and simplify,

z2 dv - 4z dv + vz dz - 4v dz + 4z dv + dv = 0

(z2 + 1) dv + v(z - 4) dz = 0

dvv

+ z−4

z2+1 dz = 0

Integrate to get,

ln v + 12

ln (z2 + 1 ) - 4 arc tan z = c

Substitute z = u/v,

ln v + 12

ln (u2/v2 + 1 ) - 4 arc tan u/v = c

Finally, substitute u = x - 1 and v = y + 2

to get

ln { (x – 1)2 + (y + 2 )2} - 8 arc tan x−1y+2

= c

75

Exercise No. 12

D.E. Linear in x and y


1. (2x – 3y + 2)dx – (3x + 2y + 1) dy = 0

2. (x + 2y + 1) dx + (2x + 4y – 3) dy = 0

76

3. (2x – 3y – 4) dx – (3x – 4y – 2) dy = 0

4. (x – 2y – 7) dx + (2x – 4y + 1) dy = 0

5. (2x – y – 3) dx – (x + 4y + 3) dy = 0

77

6. (2x – y + 1) dx + (x – 2y – 1) dy = 0

7. (x – y – 6) dy = (x – y + 2) dx

8. (x – 2y + 5) dx + (x – 2y – 1) dy = 0

78

9. (x – 2y + 4) dx + (2x – 4y – 1) dy = 0

10. (5x + 3y – 4) dx + (x + y – 2) dy = 0

79

Unit 4

Higher-Ordered Differential Equations

4.1 SPECIAL SECOND-ORDERED DIFFERENTIAL EQUATIONS

CASE 1: d 2 y = f(x) dx2

Solution: y = x ∫ f(x) dx - ∫ x f(x) dx + c1x + c2

CASE 2: d 2 y = g(y) dx2

Solution: x = . dy . + c2

2 ∫ g(y) dy + c1

CASE 3: d 2 y = f dy dx2 dx

Solution: x = ∫ dp + c f(p)

or y = ∫ pdp + c f(p)

where: p = dy dx

CASE 4: d 2 y = f x, dy dx2 dx

Solution: Let p = dy and dp = d 2 y dx dx dx2

CASE 5: d 2 y = f y, dy dx2 dx

Solution: Let p = dy and p dp = d 2 y dx dy dx2

CASE 6: d { dy + y P(x) } = Q(x)

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dx dx

Solution: Let u = dy + y P(x) dx

REMARK: Cases 4 to 6 will reduce to first-ordered D.E. upon the application of the suggested substitution. Appropriate method/s is/are applied. (Methods 1 to 7, under Section 3)

Example 4-1 y” = x2 - 1

Solution: Under CASE 1, the solution of this D.E. is

y = x ∫ f(x) dx - ∫ x f(x) dx + c1x + c2

where: f(x) = x2 - 1

Substitute to the D.E.,

y = x ∫ (x2 - 1)dx - ∫ x (x2 - 1) dx + c1x + c2

Integrate to give,

y = 1 x4 - 1 x2 + c1 x + c2

12 2

Example 4-2 y” = y + 2

Solution: Under Case 2, the formula is

x = . dy . + c2

2 ∫ g(y) dy + c1

where: g(y) = y + 2 and

∫ g(y)dy = y 2 + 2y 2

Substitute to the formula

x = . dy . + c2

(y2 + 4y + c1)1/2

Use the trigonometric substitution, thus

x = ln{2y + 4 + 2(y2 + 4y + c1)1/2} + c2

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Example 4-3 y” = 3y’Solution This 2nd-ordered D.E. is under Case 3

where f(p) = 3y’ or 3 dydx

and p = dy dx

Use the formula,

x = ∫ dp + c1

f(p)x = ∫ dp + c1

3pand x = 1 ln p + c1

3In the exponential form, this is reducible to

p = c1e3x = dy dx

Variable separable method is applied,

dy = c1e3xdx Finally,

y = 1 c1e3x + c2 Ans. 3

Example 4-4 xy” – 2x(y’)2 + y’ = 0

Solution: The D.E. is under Case 4 where the variable y is missing.

Let p = dy or y’ dx

dp = d2y or y”dx

Substitute to the D.E.

x dp – 2x(p)2 + p = 0 dx

or dp + p = 2p2 (Bernoulli equation)

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dxHere, P = 1 ; Q = 2; n = 2

x

The solution is obtained to be

p-1x-1 = c1 – 2 ln x

and p = dy = . 1 . dx x (c1 - 2ln x)

Separation of variables finally gives,y = - 1 ln(c1 - 2ln x) + c2

2

Example 4-5 yy” + 2(y’)2 = 0

Solution: This is under Case 4 where the variable x is missing.

Let p = y’ and p dp = y” dy

and substitute to the given D.E.,

y p dp + 2 p2 = 0 dy

or y dp + 2p = 0dy

Separation of variables and integration gives

ln p + 2 lny = c

or ln p + ln y2 = c

ln py2 = c

py2 = ec = c1

p = c1y-2

Since p = dy = c1y-2

dx

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c1 dx = y2 dy

Finally, c1x = y 3 + c2 Ans. 3

Example 4-5 d (dy/dx + y) = e-x

dx

Solution: This 2nd –ordered D.E. falls under Case 6Let u = dy + y

dx

and so, du = e-x

dx

or u = -e-x + c1

Substitute back to u = dy + y dx

This is FOLDE where

P = 1 ;Q = -e-x + c1

and = e∫ P dx = ex

The solution is

y = ∫ Q dx + c2

y ex = ∫ (-e-x + c1 ) ex dx + c2

y ex = -x + c1 ex + c2

Finally,

y = c1 + c2 e-x - x e-x

Exercise No. 13

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Special Second-Ordered D.E.

Name: ____________________________ Score: ___________ Rating: ___________. Find the general solution of the following differential equations.

1. y” + y = 02. (x2 – 1)y” + xy” = 03. y” + 2y = 04. y” + y’ = 05. y” – (y’)2 - 1 = 06. y” + yy’ = 07. yy” - 2(y’)2 + y2 = 0

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4.2 HOMOGENEOUS HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS (HHOLDE)

1.The general form of the HHOLDE is:

Normal Form:

A0 d n y + a1 d n–1 y + . . . + an –1 dy + any = 0 dxn dxn-1 dx

Operator Form

(a0Dn + a1Dn-1 + . . . +an-1 D + an)y = 0

where: D = d ; Dr = d r (r 0) dx dxr

The coefficients a’s are generally functions of x.

2.Connected with the HHOLDE is an “auxiliary” equation having the form,

a0mn + a1mn-1 + . . . + an-1m + an = 0

where m which replaced D, is called the auxiliary or characteristic equation. The form of the solution of HHOLDE depends upon the nature of the roots of the auxiliary equation.

Four cases arises:

Case 1. The auxiliary equation has real and distinct roots. None of which is repeated. If the real and distinct roots of the auxiliary equation are m1, m2, . . . mn-1, and mn, then the solution of the HHOLDE is,

y = c1em1x + c2em2x + . . . + cnem n x

Case 2. The auxiliary equation has real and repeated roots. Suppose a root m i of the auxiliary equation is real and multiplicity p, then the part of the solution of HHOLDE for this repeated root is

y = (c1 + c2x + c3x2 + . . . cpxp-1) emi x

Case 3. The auxiliary equation has some complex or imaginary roots, none of which is repeated. If all the a’s are real constants, then the complex

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roots of the auxiliary equation will occur in conjugate pairs. Let a bi be such a pair. Then the part of the solution of HHOLDE for this pair is

y = ea x (c1 cos bx + c2 sin bx)

Case 4. The auxiliary equation has some repeated complex or imaginary roots. If the conjugate pair a bi is a root of the auxiliary equation repeated p times, then the part of the solution of HHOLDE for these roots is

y = ea x [(c1 + c2x + . . . + cpxp-1) cos bx + (cp + 1 + cp+2 x + . . .+ c2p xp-1) sin bx]

REMARK: If the nature of the roots of the auxiliary equations happen to be a combination of real and complex numbers, the above cases may be collectively applied.

Examples for Case 1: All roots real and distinct.

Example 5-1 (D3 – 7D2 + 12D)y = 0.

Solution: The auxiliary equation is

m3 – 7m2 + 12m = m(m – 3)(m – 4) = 0

m = 0, 3, and 4

y = c1 + c2e3x + c3e4x . Answer.

Example 5-2 (D3 - 5D2 + 2D + 8)y = 0.


m3 - 5m2 + 2m + 8 = 0

By synthetic division the roots are found to be

M = 2, 4, and –1

The solution is: y = c1e2x + c2e4x + c3e-x Answer.

Examples for Case 2: Auxiliary equation has repeated real roots.

Example 5-3 (D2 - 4D + 4)y = 0.

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m2 - 4m + 4 = 0 ; (m – 2)2 = 0

m = 2 (twice)

y = (c1 + c2x)e2x . Answer.

Example 5-4 (D2 - 6D + 9)2 y = 0.


(m2 - 6m + 9)2 = 0 ; (m – 3)4 = 0

m = 3 (4 times)

y = (c1 + c2x + c3x2 + c4 x3) e3x . Answer.

Examples for Case 3: Auxiliary equation has some complex roots, none repeated.

Example 5-5 (D2 – 8D + 25)y = 0

Solution: m2 – 8m + 25 = 0 ; m = 4 3i.

y = e4x(c1 cos 3x + c2sin 3x). Answer.

Example 5-6 (D3 – 8D2 + 25 D)y = 0

Solution: m3 - 8m2 + 25m = 0 ; m(m2 – 8m + 25) = 0

m = 0 , 4 3i

y = c1 + e4x(c2 cos 3x + c3sin 3x). Answer.

Example 5-7 (2D4 + D3 - 4D2 – 10D - 4)y = 0

Solution: 2m4 + m3 - 4m2 – 10m - 4 = 0

By synthetic division, we find the roots to be

m = 2, - ½ , and -1 I

y = c1e2x + c2e-x/2 + e-x (c3 cos x + c4 sin x). Answer

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Examples for Case 4: Auxiliary equation has complex roots, some repeated.Example 5-8 (D2 + 36)2y = 0

Solution: (m2 + 36)2 = 0 ; m = 6i (twice)

y= (c1 + c2x) cos 6x + (c3 + c4x) sin 6x. Answer.

Example 5-9 (D2 + 4D + 20)2y = 0

Solution: (m2 + 4m + 20)2 = 0; m = -2 4i (twice).

y = e-2x [(c1 + c2x) cos 4x + (c3 + c4x) sin 4x] Answer.

Exercise No. 14

Homogeneous Higher-Ordered Linear D.E.



1. (D3 + 2D2 – D – 2)y = 0 Answer: y = c1ex + c2e-x + c3e-2x

2. D2(D + 2)4y = 0 Answer: y = c1 + c2x + e-2x(c3 + c4x + c5x2 + c6x3)

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3. (D3 + 4D2 + D – 6)y = 0 Answer: y = c1ex + c2e-2x + c3e-3x

4. (D3 + 6D2 + 25D)y = 0 Answer: y = c1 + e-3x(c2cos 4x + c3 sin 4x)

5. (D2 – 2D – 3)3y = 0 Answer: y = (c1 + c2x + c3x2)e3x + (c4 + c5x + c6x2)e-x

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6. (D2 – 10D + 34)3y = 0 Answer: y = e5x[(c1 + c2x + c3x2)cos 3x + (c4 + c5x + c6x2)sin 3x]

7. (D – 2)2 (D + 3)4y = 0 Answer: y = (c1 + c2x)e2x + (c3 + c4x + c5x2 + c6x3 )e-3x

8. (D2 + 12D + 40)4y = 0 Answer: y = e-6x[(c1 + c2x + c3x2

+c4x3)cos 2x + (c5 + c6x + c7x2 + c8x3)sin 2x]

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9. (D2 – 3D – 4)3y = 0 Answer: y = (c1 + c2x + c3x2)e4x + (c4 + c5x + c6x2)e-x

10. (D2 – 2D – 3)3(2D2 – 5D – 7)2y = 0 Answer: y = (c1 + c2x + c3x2)e3x

+ (c4 + c5x + c6x2 + c7x3 + c8x4)e-x

(c9 + c10x)e7x/2

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.4.3 NON-HOMOGENEOUS HIGHER ORDER LINEAR DIFFERENTIAL EQUATIONS (NHOLDE)

4.3.1 NHOLDE with Constant Right-Hand Function

a. Operator Form:

(a0Dn + a1Dn-1 + . . . +an-1 D + an)y = f(x) 0

where: D = d ; Dr = d r (r 0) dx dxr

b. Solution of the NHOLDE with Constant a’s and f(x) = Fo (a constant)

CASE 1: When an 0

Solution: y = yc + yp

where: yp = F0

an

yc is obtained by the method of Section 5

CASE 2: when an = 0

Solution: y = yc + yp

where: yp = F0 x k k!an-k

(k = order of the lowest ordered derivative)

yc is obtained by the method of Section 5

REMARK: For the case where the right-hand function f(x) is not a constant, three methods of solution are commonly used:

(1) Reduction of Order; (2) Undetermined Coefficients; (3) Variation of Parameters. A special higher order D.E. with variable coefficients, the Euler equation will be discussed in example problems – together with the above-mentioned

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methods of solving a non-homogeneous higher linear D.E. (NHOLDE) with f(x) constant.

NOTE: In the solutions under section 6.1, yc and yp are known as the “complementary function” and “particular integral”, respectively. yp, being a particular solution, does not contain any arbitrary constants.

Example 6-1 Find the complete solutions of the following:

(a) (D2 + 4D + 4)y = 12

(b) (D3 - 9D)y = - 27

Solution :

(a) (D2 + 4D + 4)y = 12 is “non-homogeneous” with f(x) = Fo = 12 and an = 4 0

(i) Complete solution: y = yp + yc

(i) For yp: Use CASE 1, SECTION 6.1

yp = Fo = 12 = 3 an 4

(ii) For yc: Auxiliary equation is,

m2 + 4m + 4 = 0

(m + 2)2 = 0

r1 = r2 = -2 (Real and equal)

By Case II, Section 5

yc = e-2x(c1 + c2x)

(iii) Finally, y = yp + yc

y = 3 + e-2x(c1 + c2x)

(b) Likewise, (D3 - 9) y = -27 is “non-homogeneous” with f(x) = Fo = -27 and an = 0.

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(i) Complete solution: y = yp + yc

(ii) For yp: Use CASE 2, SECTION 6.1

yp = Fo x k = - 27x = 3x k!an- k 1!(-9)

Here, k = 1 (order of the lowest derivative)

an- k = -9

(iii) For yc: Auxiliary equation is,

m2 – 9m = 0

m(m – 3) (m + 3) = 0

r1 = 0 ; r2 = 3 ; r3 = - 3 (Real and distinct roots)

By Case 1

yc = c1e0 + c2e3x + c3e-3x

(iv) Finally, y = yp + yc

y = 3x + c1e0 + c2e3x + c3e-3x

4.4 NHOLDE with Right-Hand Function in terms of the Variable x4.4.1 Reduction of Order Method

(1) Based form the roots of the auxiliary equation, write the given D.E as

ao(D – r1)(D – r2) . . . (D – rn)y = f(x)

and let z = (D – r2) . . . (D – rn)y

This will reduce the D.E. to

ao(D – r1)z= f(x)

which is now a FOLDE (SECTION 3.5) and can be solved in the form

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z = g(x)

(2) Substitute this back into z,

(D – r2) . . . (D – rn)y = g(x)

and let v = (D – r3) . . . (D – rn)yreducing to the form

(D – r2) v = g(x)

which is again a FOLDE and can be solved as

v = h(x)

(3) Once more substitute v to its original value and repeat the procedure until the last factor is reached, that is,

(D – rn)y = (x)

which will finally give,

y = yp = F(x)

REMARK: Since yp is a “particular” solution, the constant of integration is omitted in the procedure outlined above. The value of yc to complete the solution by using the theorems on homogeneous D.E.

Example 6-3 Find the complete solution of: (D2 – 4D + 4) y = ex usingreduction of order method.

Solution:

General Procedure:

(i) In the given D.E., (D2 – 4D + 4)y = ex , the auxiliary equation is,

m2 – 4m + 4 = 0(m – 2)2 = 0

r1 = r2 = 2 (real and equal)

So, by Case 2, yc = e2x (c1 + c2x)

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(ii) To get yp by the Reduction of Ordre method, write the given D.E. in the factored form,

(D – 2) (D – 2)y = ex

Let z = (D – 2)y, and so

(D – 2)z = ex

or dz - 2 z = ex (FOLDE) dx

with P = -2 ; Q = ex ; = e-2x

So, by the formula,

z = ∫ Q dx

z e-2x = ∫ ex e-2x dx

z e-2x = - e-x

z = - ex

Now, (D – 2)y = z = - ex

or dy - 2y = - ex (FOLDE)

with P = -2; Q = -ex ; = e-2x .

Finally, y = ∫ Q dx

y e-2x = ∫ (-ex) e-2x dx

y e-2x = e-x

and y = yp = ex.

(iii) So, y = yp + yc

y = ex + e2x (c1 + c2x) Answer

Example 6-4 Solve (D2 – 6D + 8)y = 4e3x using Reduction of Order.

Solution: The complementary function is

yc = c1e2x + c2e4x

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Write the operator in the factored form:

(D – 2) (D – 4)y = 4e3x

Let (D – 4)y = z; then (D – 2)z = 4e3x or

dz - 2 z = 4e3x

dx

Solving this equation whose integrating factor is e-2x, we find z = 4e3x. There is no need to include a constant of integration since this is a part of yp. Next we solve

(D – 4)y = z = 4e3x

whose solution is yp = - 4e3x

Hence, y = yc + yp = c1e2x + c2e4x - 4e3x. Answer.

Exercise No. 15

Non-Homogeneous Higher-Ordered Linear D.E.(Reduction of Order)

Name: ____________________________ Score: _________ Rating: _________

Solve the following problems by Reduction of Order.

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1. (D2 – 1)y = 2ex Answer: y = c1ex + c2e-x + xex

2. (D2 – D – 2)y = 12x – 6e2x Answer: y = c1e2x + c2e-x – 6x + 3 - 2xe2x

3. (D2 – 2D + 1)y = e2x – x2 Answer: y = (c1 + c2x)ex + e2x – x2 – 4x - 6

4. (D2 – 5D + 4)y = 48x – 12ex Answer: y = c1ex + c2e4x + 12x +

15 + 4xex

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5. (D2 – 5D + 6)y = 15 sinx – 15 cosx Answer: y = c1e2x + c2e3x + 3 sin x

6. (D2 – D – 2)y = 4x2 Answer: y = c1e-x + c2e2x – 2x2 + 2x – 3

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7. (D2 – 4D + 3)y = 20 cosx Answer: y = c1ex + c2e3x + 2 cos x– 4 sinx

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8. (D2 – 1)y = 10 sin2x Answer: y = c1ex + c2e-x + cos 2x - 5

9. (D2 – 1)y = 10 e2xsinx Answer: y = c1ex + c2e-x + e2x( sinx – 2 cos x)

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10. (D2 + 4D + 4)y = 12x2 e-2x Answer: y = (c1 + c2x)e-2x + x3e–2x

4.4.2 Undetermined Coefficients Method

This method for finding yp has a decided advantage over the first, since it requires no integration processes but ony differentiation. However, it can be applied only to special types of the right-hand function, listed under the following rules:

RULE 1: When f(x) = K xpeqx ( q = 0; an 0)

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yp = Axp + Bxp – 1 + . . . + Lx + M

RULE 2: When f(x) = K xpeqx (q = 0; an = 0)

yp = (Axp + Bxp – 1 + . . . + Lx + M) xr

where: r is the number of times 0 appears as a root of the auxiliary equation.

RULE 3: When f(x) = Kxp eqx (q 0)

yp = eqx (Axp + Bxp – 1 + . . . + Lx + M) xr

where: r is the number of times q appears as a root of

the auxiliary equation.RULE 3: When f(x) = Kxp eqx cos bx or Kxp eqx sin bx

yp = eqx (A1 xp + B1 xp – 1 + . . . + L1 x + M1 ) xr cos bx eqx (A2 xp + B2 xp – 1 + . . . + L2 x + M2 ) xr sin bx

where: r is the number of times q + bi appears as a root of the auxiliary equation.

Example 6-5 In the given D.E., (D2 – 4D + 4)y = e , we have

4.4.3 Variation of Parameters Method 4.5 The Euler Equation

Unit 5

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GEOMETRICAL AND PHYSICAL APPLICATIONS OF ORDINARY DIFFERENTIAL EQUATIONS

1.1 Geometrical Applications

1.1.1 Rectangular Curves

Example 7-1 Find the equation of the family of curves for which the slope at any point (x, y) is x .

y

Solution: The slope at any point is equal to the derivative.

dy = x or xdx = ydydx y

Integrating, x2 = y2 + C1 or

x2 - y2 = c2 , (c2 = 2c1). Answer.

Example 7-2 Find the equation of the family of curves for which the slope at any point equals of the product of its coordinates.

Solution:

In this case, dy = xy or dy = xdx dx dy

Integrating, ln y = x 2 + k and 2

2

y = ek ex /2 2

y = c ex /2 Answer.

1.1.2 Orthogonal Trajectories

An orthogonal trajectory of a given family of curves is a curve that is perpendicular to every curve of the family. Thus, if dy/dx is the slope of the given family of curves at any point, then for the orthogonal trajectories,

dy = - 1 .dx O.T. dy

dx

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Example 7-3 Find the equation of the system of orthogonal trajectories of the parabolas x + y = a, a > 0.Solution: From x + y = a, we have

1 + dy/dx = 0 or dy = - y . 2x 2y dx x

For the orthogonal trajectories,

dy = + x . or x dx = y dy.dx y

Integrating, 2 x3/2 _ 2 y3/2 = c or x3/2 _ y3/2 = c Answer. 3 3

Example 7-4 Find the equation of the system of orthogonal trajectories of the family of circles x2 + y2 = cx.

Solution: Differentiating the given equation, we get,

x2 + y2 = cx , 2x + 2yy’ = c

x2 + y2 = (2x + 2yy’) x = 2x2 + 2xyy’

y’ = slope of the circles at any point

= y 2 - x 2 2xy


dy = _ 2xy = 2xy .dx y2 – x2 x2 – y2

2xy = (x2 – y2)

2xydx - x 2 dy + dy = 0 ; y2

x 2 + y = k ; x2 + y2 = ky. Answer. y

To sketch the given family of curves and the orthogonal trajectories, transform the equation to their standard forms. The standard forms are:

O.T.

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For the given family of curves:

(x – c )2 + y2 = c 2 2 4For the orthogonal trajectories:

x2 + (y – k )2 = k 2 a. 4

Example 7-5 Find the equation of the system of orthogonal trajectories of the ellipses 2x2 + y2 = c.

Solution:2x2 + y2 = c , 4x + 2yy’ = 0 , y’ = -2x

y


dy = + y ; 2 dy = dx

2lny = ln x + lnk

ln y2 = lnkx

y2 = kx Answer.

Thus, the orthogonal trajectories are the parabolas y2 = kx, where k is an arbitrary constant.

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1.1.3 Isogonal Trajectories

1.2 Physical Applications of Ordinary Differential Equations

1.2.1 Law of Exponential Change

In the physical world, there are elements/objects which continually undergo changes. Bacteria, for example, increases its count with time, while radioactive elements decreases its “strength” or energy as time increases. It may be possible that the rate at which the amount of this element or object is proportional (or ONLY approximately proportional) to the amount at any instant. In which case, the complete behavior of the

element or object as it undergoes change may be defined by an exponential function – in the following manner:

(2) From the above stated rate of change,

dA = k A dt

where: A is the amount of the element or object at any instant.(3) Separation of variables and integration gives,

Ln A = kt + C

or A = ekt + C = ekteC

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Finally, A = C ekt

NOTE: (ii) For the basic equation, A = C ekt , to be completely defined, two (2)

boundary conditions are necessary for the determination of the arbitrary constants C and k.

(iii) From the resulting basic equation, the law that defines this process is termed “law of exponential change.”

(iv) When k > 0: Change is known as “growth”. When k < 0: Change is called “decay”.

Example 7-6: A certain radioactive element follows the “law of exponential change” and has a “half-life” of 38 hours. Find (a) how long it takes for 90% of the radioactivity of the element to be dissipated; (b) the percentage of radioactivity that remains after 76 hours. (NOTE: “Half-life” of an element is the time required for 50% of the radioactivity to be dissipated.)

Solution:

Let RP = percentage of radioactivity of the element at any instant

Basic Equation: RP = C ekt.

Given: When t = 0 (present time); RP = 100% When t = 38 hrs (half-life); RP = 50%

Required: When t = ? ; RP = 10%When t = 76 hrs; RP = ?

From the GIVEN conditions,

(1) 100% = Ce0 and C = 100%(2) 50% = C e38k = 100% e38j

and e38k = 0.5038k = - 0.693

k = - 0.018

So, RP = 100% e-0.018t

(4) For RP = 10% (90% dissipation),

10% = 10% e-0.018t

e-0.018t = 0.10-0.018t = - 2.30

t = 128 hours. Answer.

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(5) For t = 76 hours,RP = 100% e-0.018(76)

RP = 25.5% Answer.

Example 7-7: A bacterial population follows the “law of exponential change.” If between noon and 2:00 P.M. the population triples, at what time would the population become 100 times what it was at noon? At 10:00 A.M., what was the percentage of the bacterial population in terms of the population at noon?

Solution: Let Pb = bacterial population at any instant

Basic Equation: Pb = C ekt

Given: When t = 0 (NOON); Pb = Po

When t = 2 hrs (2 P.M.); Pb = 3Po

Required: When t = ?; Pb = 100Po

(3) From the GIVEN conditions,(1) Po = C eo or C = Po

(6) 3Po = C e2k = Po e2k

or e2k = 32k = 1.10

and k = 0.55So, Pb = Poe0.55t

(4) (1) When Pb = 100Po,100Po = Po e0.55t

e0.55t = 1000.55t = 4.605

t = 8.37 hrs

Time is: 8:22.2 P.M. Answer(3) When t = -2 hours (10 A.M.)

Pb = Poe0.55(-2) = 0.33 Po

So, at 10:00 A.M., the bacterial population was 33% of the population at noon.

Example 7-8: The rate of change of the male population of a particular barangay is proportional to the male population at any time. If the present male population of the barangay is 30,000 and ten (10) years ago, there were 20,000 males in the barangay, when will the population double its present count? How many males were there 4 years ago?

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Solution: Let Pm = male population of the barangay at any instant

Basic Equation: Pb = C ekt

Given: When t = 0, Pm = 30,000When t = - 10 years; Pm = 20,000

Required: When t = ? ; Pm = 60,000When t = -4 years; Pm = ?

From the GIVEN conditions,

(1) 30,000 = Ce0 or C = 30,000

(2) 20,000 = C e-10k = 30,000 e-10k

e-10k = 0.667– 10k = - 0.405 k = 0.405

So, Pm = 30,000 e0.0405

(1) When Pm = 60,00060,000 = 30,000 e0.0405t

e0.0405t = 20.0405t = 0.693

t = 17.10 years

(2) When t = - 4 years,Pm = 30,000 e0.0405(-4)

Pm = 25,514Exercise No. 20

Law of Exponential Change

Name: ___________________________ Score: _________ Rating: _________

Solve the following problems.

1. Radium decomposes at a rate proportional to the quantity present. If at the end of 20 years approximately 1% of the original quantity has decomposed, how long will it take for one-half of the original quantity to decompose? (Answer: 1,379 years)

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2. A certain radioactive substance has a half-life of 50 hours, that is, half of the original quantity has decomposed at the end of 50 hours. How long will it take for 95% of the substance to dissipate? (Answer: 216 hours)

3. If I represents the intensity of light which has penetrated to a depth of x meters, then the arte of change of I is directly proportional to I. If I at a rate of 10 m is 4/9

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of the intensity at the surface, find the intensities at depths of 20 meters and 40 meters.(Answers: 0.1975 Io, 0.0390 Io)

4. The rate of population growth of a certain country is proportional to the number of inhabitants. If the population of a certain country is now 4o million and 50 million 10 years later, what will be its population 20 years from now?(Answer: 62.5 million)

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5. The rate at which a chemical substance decomposes is proportional to the amount of the substance still unchanged. What is the proportional constant if the amount of the substance changes from 1000 grams to 500 grams in 2 hours? (Answer: - 0.347)

6. A certain radioactive element changes at a rate proportional to the amount at any instant (law of exponential change). How many grams of the element will be left in 300 years, if 100 grams was set aside 50 years ago and at present is only 90 grams?(Answer: 48 grams)

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7. A certain radioactive element dissipates at a rate proportional too its amount instantaneously present. If one-half of any given amount of the element dissipates in 1,600 years, what fraction will dissipate during the second century?(Answer: 8.3% or about 2/25 part)

8. A certain spherical object increases in mass at a rate proportional to the instantaneous surface area of the object. If half its mass was gained in 100 seconds, how long will its take its radius to increases to twice its initial value. In 200 seconds, what percentage increase of its radius was gained by the object? (Answer: 385 seconds; 52%)

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1.2.2 Law of Newton on Cooling

The Newton’s Law on Cooling” is stated as follows: “The rate of change of the temperature of a body placed in a medium of constant temperature is proportional to the difference between the temperatures of the body and medium – provided this differences is NOT large.” In differential equation form

dTb = k(Tb - Tm) dt

which when solved by separation of variables,

Tb = C ekt + Tm

where: Tb = variable temperature of the body Tm = constant temperature of the medium

Example 7-9 The temperature of a metal is 20C. It is submerged in a large body of liquid of temperature 5C. After one (1) minute, the temperature of the metal dropped down to 10C. When will the metal’s temperature be 6C? After 30 seconds, what will be temperature of the metal?

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Solution: Basic Equation: Tb = Cekt + Tm

Where: Tm = 5C (liquid medium)

GIVEN: When t = 0; Tb = 20C

When t = 1 min; Tb = 10C

REQUIRED: When t =? ; Tb = 6C When t = 0.50 min ; Tb = ?

(i) From the GIVEN conditions,

(1) 20 = C e0 + 5 or C = 15

(2) 10 = C ek + 5 = 15 ek + 5

or ek = 5 15 k = - 1.10

So, Tb = 15 e-1.10t + 5

(ii) (1) When Tb = 6C,

6 = 15 e-1.10t + 5

e-1.10t = 1 15 - 1.10t = - 2.71 t = 2.46 min Answer

(2) When t = 0.50 min (30 seconds)

Tb = 15 e-1.10(0.50) + 5

Tb = 13.65 C Answer

Example 7-10 At a certain time, a thermometer reading 70F is taken outdoors where the temperature is 15F. Five minutes later, the thermometer reading is 45F. After another five minutes, the thermometer is taken back indoors where the temperature is fixed at 70F. What is the thermometer reading ten minutes after it was brought back

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indoors? When will the reading, to the nearest degree, return back to its original reading of 70F?

Solution:

Basic Equation: Tb = Cekt + Tm

Where: Tm = 15F (gas medium)

GIVEN: When t = 0; Tb = 70F

When t = 5 min; Tb = 45F

REQUIRED: Tb = ? after 10 minutes outdoors and another 10 minutes indoor t = ? when Tb = 69F (indoor)

(i) From the GIVEN conditions,

(1) 70 = C e0 + 15 or C = 55

(2) 45 = C e5k + 15 = 15 e5k + 15

or e5k = 30 55 5k = - 0.606 k = - 0.1212

So, Tb = 55 e-0.1212t + 15

NOTE: This equation is true for the case where the thermometer is outdoors.

(ii) Five (5) minutes further, the thermometer was taken back indoors. The reading here is,

Tb = 55 e-0.1212t + 15

Tb = 31.4F

So, in bringing back the thermometer indoors,

Tb = 31.4F when t = 0

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and the temperature of the medium, Tm = 70F, resulting into a new relation,

Tb = C ekt + 70

Substitute Tb = 31.4F when t = 0

31.4 = C e0 + 70 or C = -38.6

and so, Tb = -38.6 e-0.1212t + 70

(iii) When t = 10 min (indoors),

Tb = -38.6 e-0.1212(10) + 70

Tb = 58.5F Answer

(iv) When Tb = 69F

69 = -38.6 e-0.1212t + 70

e-0.1212t = 1 . 38.6

t = 30 minutes

NOTE: The value of k for the case when the thermometer is inside is the same as when it is outside.

Exercise No. 21

Newton’s Law on Cooling


1. If in air at 60 C a body cools from 90 C to 80C in 10 minutes, find its temperature 20 minutes later. (Answer: 73.35C)

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2. A thermometer reading 50 C is taken outside where the temperature is 30 C. Five minutes later the reading is 40C. Find the temperature reading 10 minutes later after the thermometer was brought outside.(Answer: 35C)

3. In Problem 2, find the time it will take for the temperature to drop from 50 C to within 30.5 C. (Answer: 26.62 minutes)

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4. A 12:00 noon a thermometer reading 60 C is taken outside where the air temperature is 24 C. At 12:02 P.M. the reading is 32 C. At 12:06 P.M. the thermometer is taken back indoors where the temperature is 60C. What is the thermometer reading at 12:10 P.M.? (Answer: 58.24 C)

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5. A thermometer reading 10C is brought into a room where the temperature is 32C. Two minutes later the thermometer reading is 15. Determine the temperature reading 5 minutes after the thermometer is first brought into the room. (Answer: 20.26C)

6. A thermometer reading 45C is brought outside where the temperature is 25C. Five minutes later the thermometer is 33C. Find (a) the thermometer reading 10 minutes after the thermometer was brought outside; and (b) the time it will take for the reading to drop to 26C.(Answer: 28.2C)

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7. At 1:00 P.M. a thermometer reading 35C is taken outside there the air temperature is 6 C. At 1:04 P.M. the reading is 18 C. At 1:09 P.M. the thermometer is taken back indoors where the temperature is 35 C. What is the thermometer reading at 1:15 P.M. ? (Answer: 29.84C)

8. The rate at which a body is cooling to the difference in the temperatures of the body and the surrounding medium. If a body cools from 120 to 70F in one (1) hour, after placing it in a medium whose temperature is 20F, how long will the body cool to 30F? After 2 hours, what will be the temperature of the body? (Answer: 45 F, 3 hours, 19 minutes)

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1.2.3 Simple Electric Circuits

(1) A simple electric circuit (SEC) is one in which an inductance L ( henry), a resistance R (ohm), and a capacitance C 9farad), are connected in series with a source of electromotive force, EMF of E volt). The differential equation necessary to solve the various problems relative to this circuit is derived with the use of the Ohm and Kirchhoff’s laws.

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(2) The first law of Kirchhoff states that “the algebraic sum of all the potentials around a closed electric circuit is zero.” Applying in this law to the SEC being considered, we have

E – EL – ER – Ec = 0

where: EL = L dI (volt) dt

ER = RI (volt)

EC = Q (volt) C

So, L dI + RI + Q = E dt C

and L d 2 Q + RdQ + Q = E dt2 dt C

NOTE: (i) The two (2) basic equations used in SEC are:

(1) When the capacitance is not present,

L dI + RI = E dt

(2) With the capacitance in the SEC,

L d 2 Q + RdQ + Q = E dt2 dt C

or ( LD2 + RD + 1 ) Q = E C

With D = d dt

(ii) The inductance L is a circuit parameter which opposes a change in the current I (ampere) causing a potential drop of EL.

(iii) The resistance R is a circuit parameter which opposes the current I causing a potential drop ER.

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(iv) The capacitance C is a circuit parameter which opposes a change in the voltage causing a potential drop EC.

(v) The current I is the rate of change of the electric chanrge Q(coulomb) or

I = dQ dt

and so dI = d 2 Q dt dt2

Example 7-11 A coil of inductance 1 henry and resistance 10 ohms is connected in series with an EMF of Eo sin 10t volts. When t = 0, the current is zero. If I = 5 amperes when t = 0.1 second, what must be the value of Eo?

Solution: This is a SEC where there is NO capacitance, and so the basic equation is,

L dI + RI = E dt

where: L = 1 henry

R = 10 ohms

E = Eosin 10t volts

(i) Substitute all known values,

1 dI + 10 I = E osin 10t (FOLDE) dt

with P = 10; Q = E osin 10t ; = e10t

(ii) Solution is: I = ∫ Q dt + C

I e10t = ∫ E osin 10t e10t dt + C

Evaluate the integral to get,

I e10t = E o ( e 10t ) (10 sin 10t – 10 cos 10t) + C 200

or I = E o (sin 10t – cos 10t) + C e-10t

20

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But I = 0 when t = 0, giving C = Eo 20

So, I = E o (sin 10t – cos 10t + e-10t) 20

(iii) Finally, for t = 0.1 second, I = 5 amperes,

5 = E o (sin 1 – cos 1 + e-1) 20

giving Eo = 149 volts

Example 7-12 A sinusoidal EMF 110 sin 400t volts is connected in series in a circuit with an inductance of 0.1 henry, a resistance of 10 ohms, and a capacitance of 250 microfarad. Find the steady-state solutions of the charge Q and the current I in terms of the time t. Also determine the maximum values of the steady-state Q and I.

Solution: This is SEC contains all the parameters L, R, and C.

Basic Equation: ( LD2 + RD + 1 ) Q = E C

where: L = 0.10 henry

R = 10 ohmsC = 250 x 10-6 farad

E = 110 sin 400t volts

(i) Substitute all known values,

(0.10D2 + 10D + 4000)Q = 110 sin 400t

NOTE: The “steady-state” solutions for this SEC with “sinusoidal” EMF are:

(1) Qss = _ Eo (X sin t + R cos t) Z2

(2) Iss = Eo (R sin t - X cos t) Z2

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With = 400

X = L - 1 = 30 (“reactance”) C

Z2 = R2 + X2 = 1000 (‘impedance”)

(ii) With the computed value, we have

Qss = - 0.00275 (3 sin 400t + cos 400t) AnswerandIss = 1.10(sin 400t – 3 cos 400t) Answer

(iii) To dtermine the maximum values of Qss and Iss sset their derivatives rto zero, solve for the time t and substitute back to the basic equations of Qss and Iss.

(1) dQss = _ 0.00275 (1200 cos 400t – 400 sin 400t) = 0

or 400 t = 1.25 radians

So, max. Qss = -0.00275( 3 sin 1.25 + cos 1.25)

= 0.0087 coulomb Answer

(2) dIss = 1.10 (400 cos 400t + 1200 sin 400t) = 0

or 400t = - 0.32 radian

So, max. Iss = 1.10( - sin 0.32 - 3 cos 1.25) = 3.48 amperes Answer

Example 7-13 A coil of inductance 1 henry and negligible resistance is connected in series with a capacitance of 10-6 farad and an emf E volts. Considering Q and I are both zero when t = 0, find the charge Q and current I when t = 0.001 second with (a) E = 100 volts; (b) E = 100 sin 500t volts.

Solution: The basic equation here is (LD2 + RD + 1 )Q = E C

Where: L = 1 henry

128

R = 0 (negligible)

C = 10-6 farad

E = 100 volts; 100 sin 500t volts

(i) Substitute given values (for E = 100 volts),

(D2 + 106) Q = 100 (NHOLDE with Constant Right-Hand Function)

(ii) (1) Auxiliary equation: m2 + 106 = 0

with roots: r1 = 1000i ; r2 = -1000i

so, Qc = eo(C1cos 1000t + C2 sin 1000t)

(2) Particular integral, Qp = 100 = 10-4

106

So, complte solution is: Q = Qc + Qp

Q = C1 cos 1000t + C2 sin 1000t + 10-4

and I = dQ = - 1000C1 sin 1000t + 1000C2 cos 1000t dt

(iii) Values of C1 and C2; Using the boundary conditions: I = Q = 0

When t = 0, which gives C1 = - 10-4 and C2 = 0

Q = 10-4 (1 – cos 1000t)

and I = 0.10 sin 1000tWhen t = 0.001 second,

Q = 10-4 (1 – cos 1000t) = 4.60(10) -5 coulomb Answer

and I = 0.10 sin 1 = 8.40(10) -2 ampere Answer

(iv) For E = 100 sin 500t volts,

(D2 + 106) = 100 sin 500t

a NHOLDE with “sinusoidal” emf

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Here, Qc = C1 cos 1000t + C2 sin 1000t, which is the same as in the previous case.

The value of Qp is given by the formual

Qp = _ Eo (X sin t + R cos t) Z2 where: = 400

X = L - 1 = -1500 (“reactance”) C

Z2 = R2 + X2 = 2.25(10)6 (‘impedance”)

Eo = 100

(v) Substitute to Qp,

Qp = 1.33(10)-4 sin 500 t

and so Q = Qc + Qp

Q = C1 cos 1000t + C2 sin 1000t + 1.33(10)-4 sin 500 t

and I = dQ = 1000(-C1 sin1000t + C2 cos1000t) + 0.067 cos 500t dt

Using the boundary condition: I = Q = 0 at t = 0

To give C1 = 0 and C2 = -6.79(10)-5

(vi) Finally,

Q = - 6.70(10)-5 sin 1000t + 1.33(10)-4 sin 500t

and I = - 6.70(10)-2 cos 1000t + 0.0670 cos 500t

When t = 0.001 second

Q = - 6.70(10)-5 sin 1 + 1.33(10)-4 sin 0.50

= 7.40(10)-6 coulomb Answer

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and I = -6.70(10)-2 cos 1 + 0.0670 cos 0.50

= 2.20(10)-2 ampere

Exercise No. 22

Simple Electric Circuits


1. For the circuit in Figure 1, find the current at any instant, and show that as time goes on the current approaches the limit E/R. (Answer: I = E/R (1 – e-Rt/L)

131

2. A coil of inductance 2 henrys and a resistance of 10 ohms is connected with an emf E = 200 sin 20 t volts, where t is the time in seconds. I = 0 when t = 0. Find I when t = 0.05 second. (Answer: 2.11 amperes

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3. An inductance of 2 henrys and a variable resistance R = 4/(t + 10) are connected in series with a constant emf of E volts. If I = 0 when t = 0, determine E so that I = 50 amperes when t = 5 seconds. (Answer: 28.42 volts)

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4. A coil of inductance 1 henry and negligible is connected in series with a capacitance of 4 x 10-6 farad, and an emf of E = 200 volts. When t = 0, Q = 0 and I = 0. Find Q and I at any time t. (Answers: Q = 1/1,250(1 – cos 500t; I = 2/5 sin 500t)

134

5. When a resistance R (ohms) and a capacitance C (farads) are connected in series with an emf E (volts), the current I (amperes) is given by the equation

R dI + 1 I = dE dt C dt

If R = 1,000 ohms, C = 5 x10-4 farad, and I = 10 amperes for t = 0, find the current for t = 1 second and E = 100 volts. (Answer: 1.35 amperes)

136

6. An inductance of 1 henry, a resistance of 1,200 ohms, and a capacitance of 10 -6 farad are connected in series with an emf E = 100 sin 400t volts. If the charge and current are both zero when t = 0, find the expression for current I at any time t thereafter.[Answer: I = e-600t (0.091 (sin 8oot + 0.036 cos 800t) + 0.036 cos 800t + 0.021 sin 400t]

137

7. An inductance of 4 henrys and a resistance of 30 ohms are connected in series with an emf of E volts. If I = 0 when t = 0, determine I when t = 0.01 second if E = 220 volts. (Answer: 0.533 ampere)

138

8. A coil of inductance 1 henry and a resistance of 10 ohms is connected in series with an emf E = 100 sin 10 t volts. I = 0 when t = 0. Find I when t = 0.2 second. (Answer: 7.30 amperes)

139

9. An inductance of 2 henrys and a variable resistance R = 4/(t + 10) ohms are connected in series with an emf E volts. If I = 0 when t = 0, determine E such that I = 40 amperes when t = 5 seconds. (Answer: 22.70 volts)

140

10. An inductance of 1 henry and a resistance of 2 ohms are connected in series with an emf of Ee-t volts. When t = 0, I = 0. Determine the value of E if I = 20 amperes when t = 1 second. (Answer: 86.1 volts)

141

11. An inductance of 1 henry, a resistance of 100 ohms and a capacitance of 10-4 farad are connected in series with an emf E = 220 volts. The charge and current are both zero when t = 0. Find I when t = 0.002 second.

(Answer: 0.342 ampere)

142

12. A coil of inductance 2 henrys and negligible resistance is connected with a capacitance of 2 x 10-6 farad and an emf E = 200 volts. If the charge Q and the current I are both zero when t = 0, find Q and I when t = 0.001 second.(Answer: Q = 0.000 0488 coulomb; I = 0.094 ampere)

144

13. An inductance of 1 henry, a resistance of 3,000 ohms, and a capacitance of 5 x 10-7

farad are connected in series with an emf of E = 200 e-1,000t volts. If the charge and current are both zero when t = 0, find I when t = 0.001 second. (Answer: 0.0194 ampere)

145

14. An inductance of 1 henry, a resistance of 1,600 ohms, and a capacitance of 10 -6 farad are connected in series with an emf of 200 sin 500t volts. Find the general expression for the charge Q. [Answer: Q = e-800t(C1cos600t +C2sin 600t) - 1/7,750 (15sin500t + 16cos500t)]

146

15. An inductance of 1 henry, a resistance of 1,200 ohms, and a capacitance of 10 -6 farad are connected in series with an emf E = 200 sin 600t volts. The charge and current are both zero when t = 0. (a) Find the value of Q and I at any time t. (b) Find I when t = 0.002 second.

1.2.4 2nd Law of Newton on Dynamics

1.2.5 A Problem on Chemical Solution1.2.6 Unsteady Flow in Orifices1.2.7 Heat Flow Problems1.2.8 Deflection of Beams Problems 1.2.9 Problems on Catenary Curves 1.2.10 Different Types of Vibrations1.2.11 Motion of a Falling Body Under a Resistance

Proportional to the Square of the Velocity1.2.12 Problems on Transmission Lines Under Steady-State Conditions1.2.13 The Tractrix

59863727 worktext in differential equations

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