differential equations - · pdf fileordinary differential equations in many physical...

DIFFERENTIAL EQUATIONS

Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats)

[email protected]

April 7, 2017

Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]


ORDINARY DIFFERENTIAL EQUATIONS

In many physical situation, equation arise which involve differentialcoefficients. For example:

1 The rate of decrease of temperature of a hot body.

2 A body falling freely under gravity.

3 The oscillation on the end of a spring.

4 The decay of a radioactive substance.

5 The decrease in concentration of chemical compound in a reaction.



FORMATION OF DIFFERENTIAL EQUATIONS

Differential equations arise or may be derived in a variety of ways. Inmost cases the problem is to find the dependent variable in terms of theindependent one. Eg. Finding x in terms of t or y in terms of x .Differential equations may be formed by direct differentiation.

Eg. y = x3 + 7x2 + 3x + 7 −→Eqn 1

dydx = 3x2 + 14x + 3 −→Eqn 2

d2ydx2 = 6x + 14 −→Eqn 3

d3ydx3 = 6 −→ Eqn 4



Definition

Equations which contain an independent variable, a dependent variableand at least one of their derivatives are called DIFFERENTIALEQUATIONS. Thus a relationship between a variable quantity x and a

dependent function y and its derivatives dydx ,

d2ydx2 ,

d3ydx3 · · · is called an

ORDINARY DIFFERENTIAL EQUATION.Equations 2,3 and 4 are differential equations.

Other examples of ordinary differential equations

1dydx = kx

2 x2(1 + y) dydx − (1 + x)y2 = 0

3d2ydx2 = −n2y



4. ( dydx )2 + 3y = 0

5. x d3ydx3 + d2y

dx2 + x( dydx )4 = 0

KINDS OF DIFFERENTIAL EQUATIONS

There are two main types of differential equatons.

Ordinary Differential Equations(O.D.E.’s) These are differentialequations with only one independent variables.

Partial Differential Equations(P.D.E’s) These are differentialequations with more than one independent variable.



Examples of partial differential equations

x ∂z∂y + y ∂z∂x = 0

∂2f∂y2 = 2x

x2+y2

∂2z∂x2 + ∂2z

∂y2 = 0

fxx + fyy = 0

∂2z∂x2 + ∂2z

∂y2 = ∂z∂t

NOTE:

This course is concerned more with ordinary differential equations.Partialdifferential equations may be considered slightly.



ORDER OF A DIFFERENTIAL EQUATION

Differential equations are classified according to the highest derivativewhich occurs in the equations.Thus if the highest derivative that occursin an equation dny

dxn ,the equation is said to be of order n.

On slide 3, we notice that:

Equation 2 is of the first order ,having only the first derivative.

Equation 3 is of the second order.

Equation 4 is of the third order.



DEGREE OF A DIFFERENTIAL EQUATION

The degree of a differential equation is the highest power of the highestderivative which the equation contains.

Thus( d2ydx2 )3 + 3 dy

dx = 0 is of the second order and third degree.Considering the examples on slide 4 and 5, we also notice that example1,2,3 and 5 are of first degree whilst example 4 is of a second degree.NOTE that in example 5 the degree of the equation is determined by the

power of the highest derivative d3ydx3 and not by the fourth power term in

dydx

Exercises

Indicate the order and the degrees of the following ordinary differentialequations.

1. ( d2ydx2 )3 + dy

dx = y − x 2. x2 dydx + x d4y

dx4 = 3y

3.( dydx )4 + xy = x 4. ex dy

dx = (1− yex)



SOLUTIONS OF A DIFFERENTIAL EQUATION

We recall that an ordinary differential equation was defined as arelationship between a variable quantity x and a dependent function yand its derivatives.These equations normally arise from physical situations and it is oftenrequired to obtain a functional relationship between x and y alone,having eliminated the derivatives. This relation is referred to theSOLUTION of the differential equation.

A solution which is COMPLETE or GENERAL must contain a number ofarbitrary constants which is equal to the order of the equation. Solutionsof the differential equation with the appropriate number of arbitraryconstants are called GENERAL SOLUTIONS.



In physical problems, solutions are usually required which satisfy certainspecified conditions. These provide information from values to beassigned to the arbitrary constants.

This type of solution , which satisfies certain definite conditions, is calleda PARTICULAR SOLUTION and the conditions satisfied are calledBOUNDARY CONDITIONS or INITIAL CONDITIONS.



Example

Considerdy

dx= x which is of the first order.

Integrating we have∫dy =

∫xdx

y =1

2x2 + A

Now if we consider the general solutions y =1

2x2 + A to the equation

dy

dx= x .

Let us assume that the boundary condition is given to be y = 1 whenx = 0⇒ A = 1

⇒ The value assigned to A = 1 and the particular solution is y =1

2x2 + 1



DIFFERENTIAL EQUATIONS OF THE FIRST ORDERAND FIRST DEGREE

Let us first look at the case where one variable is absent.

(a) When y is absent

The general form isdy

dx= f (x) ⇒

∫dy =

∫f (x)dx

y =∫f (x)dx



Example

Solve the differential equationdy

dx= x4 + sin x

⇒∫dy =

∫(x4 + sin x)dx

y =1

5x5 − cos x + c



(b) When x is absent

The general form isdy

dx= f (y)⇒ dy = f (y)dx Rewriting this in the form :

dx

dy=

1

f (y)∫dx =

∫ 1

f (y)dy =

∫ dy

f (y)

Example

Solve the equationdy

dx= tan y



Solution

dy

dx= tan y

dx

dy=

1

tan y=

cos y

sin y

x =

∫cos y

sin ydy

= ln | sin y | +c

The examples given above lead us to the main types of first order, firstdegree differential equations and their solutions.



TYPES OF FIRST ORDER DIFFERENTIAL EQUATIONS

1 Variables Separable

2 Homogeneous

3 Linear

4 Exact



TYPE 1 - VARIABLES SEPARABLE

If the terms of the equation can rearranged into two groups, eachcontaining only one variable, the variables are said to be SEPARABLE.Since the differential equations of the first order and first degree containdy

dxto the first power only, they can be written as

dy

dx= F (x , y)

In many cases F (x , y) may be written as

F (x , y) = f (x)g(y)

where f (x) and g(y) are functions of x only and g(y) is a function of yonly.



We may then ”separate the variables ” and write.

dy

g(y)= f (x)dx



Worked Example On Type 1

Example

Solvedy

dx=

5x

7y

Solution

The variables are separable

7ydy = 5xdx

⇒∫

7ydy =

∫5xdx

7

∫ydy = 5

∫xdx

7 · 1

2y2 = 5 · 1

2x2 + C ⇒ y2 =

5

7x2 + C

Note : C = C1 + C2 and C is an arbitrary constant.



Exercise 1

Solve xdx + ydy = xy(xdy − ydx)

Solution

x + ydy

dx= xy

(xdy

dx− y

)dy

dx(y − x2y) = −(x + xy2) = −x(1 + y2)∫

ydy

1 + y2= −

∫xdx

1− x2

1

2ln |1 + y2| =

1

2ln |1− x2|+ lnC1

(1 + y2)12 = C1(1 + x2)

12

(1 + y2) = C (1 + x2)Mr. Isaac Akpor Adjei(MSc. Mathematics, MSc. Biostats) [email protected]


Exercise 2

Solvedy

dx= (x + y)2

Solution

Let z = x + y

dz

dx= 1 +

dy

dx(1)

=⇒ dy

dx= z2 (2)

From (1)dy

dx=

dz

dx− 1

=⇒ dz

dx− 1 = z2



Solution Cont’d

dz

dx= z2 + 1∫

dz

z2 + 1=

∫dx

tan−1 z = x + c

z = tan(x + c)

x + y = tan(x + c)

y = tan(x + c)− x



Solve the following equations

1 idy

dx= tan2(x + y)

iidy

dx= (x + 4y)2

2 i (1 + x)y + (1− x)ydy

dx= 0

iidy

dx+

k

x2= 0

3 i (x + 1)dy − ydx = 0ii (y 2 − x2) + 2xydx



HOMOGENEOUS

TYPE 2 - - HOMOGENEOUS

M(x , y) is said to be a homogeneous function of degree n if the sum ofthe powers of x and y in each term of M is n

Eg. (i) x2y − 3xy2 + 2y3 is homogeneous of degree 3(ii) x4 − 2x2y2 is homogeneous of degree 4.

If a first order D.E. is written in the form

dy

dx=

M(x , y)

N(x , y),

where M and N are homogeneous functions of the same degree, then theequation is said to be HOMOGENEOUS.



Examples:

idy

dx=

xy

x2 + y2

ii (x2 + y2)dy

dx= xy

Check

i. (x2 + y)dy

dx= xy ?

ii.dy

dx=

y(3x2 + y2)

x(x + 3y)?



METHOD OF FINDING SOLUTION TO AHOMOGENEOUS DIFFERENTIAL EQUATION

If, in the equation;dy

dx=

M(x , y)

N(x , y), (3)

both M and N are homogeneous of degree n, we may divide them by xn

and express the R.H.S as a function of the single variable v , where v =y

x

⇒ y = vx

dy

dx= v + x

dv

dx(4)

Substituting (4) in the differential equation (3), we find that the result isa new differential equation in which the variables v and x and finally

replace v byy

x.



Example 1

(x2 + y2)dy

dx= xy

Let v =y

x

y = vx

dy

dx= v + x

dy

dx−−−−− (1)



Now (x2 + y2)dy

dx= xy

dy

dx=

xy

x2 + y2=

yx

1 + ( yx )2

dy

dx=

v

1 + v2−−−−− (2)

From (1) and (2)

v + xdv

dx=

v

1 + v2



xdv

dx=

v

1 + v2− v =

−v3

1 + v2

xdv

dx=−v3

1 + v2

Now we notice that x and v are separable. So we separate the variables,integrate and substitute for v to obtain the general solution.

Thus xdv

dx=−v3

1 + v2∫dx

x=

∫(− 1

v3− 1

v)dv

ln(x) =1

2v2− ln(v) + ln(A)

where ln(A) = C

logey

A=

x2

2y2



y

A= e

x2

2y2

y = Aex2

2y2



Example 2

Example 2

dy

dx=

y(3x2 + y2)

x(x2 + 3y2)

Let v =y

xy = vx

dy

dx= v + x

dy

dx



v + xdy

dx=

yx (3 + ( y

x )2)

1 + 3( yx )2

v + xdy

dx=

v(3 + v2)

1 + 3v2∫dx

x=

∫1 + 3v2

2v(1− v2)dv



ln(x) =1

2

∫ [1

v+

2

1− v− 2

1 + v

]dv

Finally, we have

loge x2 = loge

Av

(1− v2)2

⇒ (x2 − y2)2 = Axy



Linear Type of Ordinary Differential Equations

If a differential equation can be written in the formdy

dx+ Py = Q, where

P and Q are functions of x only, the equation is said to be LINEAR of

the first order sincedy

dxand y occurs linearly.

Examples:

1dy

dx+ 2y cot x = cos x

2dy

dx+

x

1 + x2y =

1

2x(1 + x2)

3dy

dx− x

(1− x2y =

1

(1− x2

4dy

dx+ y tan x = sec x

5dy

dx− 2xy = 2x

6dy

dx+ xy = x



METHODS OF SOLUTION

In the standard linear equation

dy

dx+ P(y) = Q,

the presence of the termsdy

dxand y suggests the differentiation of a

product involving y .To produce this product we multiply the equation throughout by afunction u to be determined later. Thus we have

udy

dx+ uP(y) = uQ



The equation reduces todu

dx= uP∫

du

u=

∫Pdx

loge u =

∫Pdx

u = e∫Pdx



We note that no arbitrary constants needs to be included here since theconstant required in the solution of the original differential equation willarise on performing the integration.The function

u = e∫Pdx

is referred to as the INTEGRATING FACTOR (I.F)



Example 1

Solve xdy

dx+ 2y = ex

Solution

xdy

dx+ 2y = ex

Writing in standard form yields

dy

dx+

2

xy =

ex

x−−−−(∗)

⇒ P =2

x

⇒∫

Pdx =

∫2

xdx = 2 ln(x) = ln(x)2



Solution Cont’d

Thus

u = e∫Pdx = e

∫2x dx

e loge(x)2

= (x)2

Multiplying (*) by u = x2, we have

x2dy

dx+ 2xy = xex

d(x2y)

dx= xex



x2y =

∫xexdx

x2y = (x − 1)ex + c1

y =ex(x − 1)

x2+

c1x2

y =ex(x − 1)

x2+ c2



Solution

Multiplying through (*) by the I.F = ln | sec x |,we have

sec xdy

dx+

sin x

cos x(sec x)y = sec x sec x

sec xdy

dx+ tan x(sec x)y = sec2 x

⇒ y · sec x =

∫sec2 x = tan x + c1

⇒ y =tan x

sec x+

C1

sec x⇒ y = sin x + c2



Exercises

Solve the following

1 (x2 + 1)dy

dx+ 2xy = 4x2 given that when

x=3, y=4

(I.F= (x2 + 1) Soln: y = 4x3

x2+1 + 4x2+1

)2 x(1− x2) dy

dx + (2x2 − 1)y = x3

(I.F = 1x√1−x2

; y=x + A1x√

1− x2 = x + A2)

3dydx = y − x(y = x + 1 + cex)

4 tan x dydx = 1 + y

(y + 1 = c sin x)



SOME APPLICATIONS OF DIFFERENTIAL EQUATIONS

We recall that a derivative is a rate of change. It is this idea that givesthe differential equation a wide range of applications in the sciences, inthe business and social sciences.Many of the applications involve a rate ofchange of some quantity with respect to time.Thus ,if the rate of change of y with respect to time , t is proportional toy, thendydt ∝ y

⇒ dydt = ky

The constant k is referred to as the constant of proportionality.Consider a general solution to a differential equation

y = cekt

If k is positive , the function represents EXPONENTIAL GROWTH



If k is negative, it represents EXPONENTIAL DECAY If t=0 ,y=cHere c is referred to as the initial value (value of y at time t=0)

SOME EXAMPLES AND EXERCISES

1 In a certain type of chemical reaction, the rate at which an oldsubstance (initial amount, a) is converted into a new substance isproportional to both the amount

2 The rate of change in temperature T of a small object placed in alarge body of water with a temperature of 32◦C is proportional tothe difference between the temperature of the object and thetemperature of the water. Find the differential equation thatrepresents this function and find its general solution.

3 Newtons Law of Cooling states that the rate of decrease oftemperature of a hot body is proportional to its excess temperatureover that of the surroundings.( dθ

dt ∝ (θ − θ0)) where θ0 is thetemperature at time t.



SOLVED EXAMPLES

Example 1

A second order rate chemical reaction is governed by the differentialequation dx

dt = k(5− x)2,where x is the change in concentration at timet. x is initially zero and is found to have the value x=1 when t=10s. Findthe value of the reaction rate constant k and the values of x when t=20sand t=100s

Solndxdt = k(5− x)2∫

dx(5−x)2 = k

∫dt = kt∫

dx(5−x)2 =?

Let u=5-x, du=-dx⇒∫

dx(5−x)2 = −

∫duu2 = −

∫u−2du = − u−1

−1 = 1u + c

= 15−x + c



⇒ kt = 15−x + c

when t=0,x=0 ⇒ c = − 15

⇒ kt = 15−x −

15

when t=10, x=110k = 1

4 −15 = 1

20

⇒ k = 1200

⇒ 1200 t = 1

5−x −15

when t=20, we have20200 = 1

5−x −15

15−x = 20

200 + 15 = 20

200 + 40200 = 60

200

5− x = 20060 = 10

3

x = 5− 103 = 5

3when t=100, we have100200 = 1

5−x −15

15−x = 1

2 + 15 = 7

10 ,⇒ 5− x = 107 ,⇒ x = 5− 10

7 = 257



CHEMICAL RATE EQUATIONS

ORDER OF A REACTION

The order of a chemical reaction is the sum of the powers of theconcentration terms that occur in the differential form of the Rateequation.Example:dxdt = k(a− x)0 ZERO ORDER REACTIONdxdt = k(a− x) FIRST ORDER REACTIONdxdt = k(a− x)2 SECOND ORDER REACTIONdxdt = k(a− x)3 THIRD ORDER REACTION

where a is initial concentration and x is the decrease in concentration inthe chemical reaction.



FIRST ORDER REACTION

A reaction in which the rate depends on two concentration terms inwhich one species is present in a very high concentration relative to theother such that its concentration considered to be constant during thecourse of the reaction results in a first order rate equationdxdt ∝ (a− x)

⇒ dxdt = k(a− x), where a is the initial concentration and x is the

decrease in concentration.

Example

1. CH3CO − O − C2H5 + H2O CH3COOH + C2H5OHdxdt = [Ester ][H2O]

2.C (CH3)3 · OH → C4H8 + H2O



Solutiondxdt = k(a− x)∫

dxa−x = k

∫dt

− ln(a− x) = kdt + c

At t=0, x=0 ⇒ c = − ln a

⇒ − ln(a− x) = kt − ln a

⇒ ln aa−x = kt −→ Eqn1

aa−x = ekt −→ Eqn1b

SECOND ORDER REACTIONS

A + B −→ CA and B are Reactants , C is the product.If x is the decrease in concentration of A at time t, and a and b are theinitial concentrations of A and B, thendxdt = k(a− x)(b − x)For special cases in which a and b are equimolar amounts.



dxdt = k(a− x)2

Thus∫

dx(a−x)2 = k

∫dt

⇒ (a− x)−1 = kt + CAt t=0, x=0, C= 1

a

⇒ 1a−x = kt + 1

a

kt = 1a−x −

1a

kt = xa(a−x) −→ Eqn2

Now let us consider the solution of the first order rate equationdxdt = k(a− x)

Solution:a

a−x = ekt

ln aa−x = kt

⇒ t = 1k ln a

a−x



Usually ln(a− x) is plotted against tk is the slopeln a is the intercept.Note: For Radioisotope work

kt = ln No

Nwhere No−initial activityand N is the activity at time t.

HALF LIFE (t 12)

The half life (t 12) is the time taken for half or 50% reaction to occur

ORthe time taken for the concentration of the reactants to reduce to half(50%) of the initial concentration.



Examples

FIRST ORDER REACTIONkt = a

a−x ⇒ t = 1t ln a

a−x

HALF LIFEAt t 1

2, x = a

2

⇒ t 12

= 1k ln a

a− a2

= 1k ln 2a

2a−a = 1k ln 2

⇒ t 12

= ln 2k = 0.6932

kThus t 1

2does not depend on a

t 12

= t50% = 1k ln a

a− 50a100

= 1k ln a

50a100

= 1k ln 100a

50a = 1k ln 2

= 0.6932k

t90% = 1k ln a

a− 90a100

= 1k ln a

10a100

= 1k ln 100

10 = 1k ln 10

= 0.1054k



t80% = 1k ln a

a− 80a100

= 1k ln a

20a100

= 1k ln 100

20 = 1k ln 5

= 0.223k

For the second order reaction,kt = x

a(a−x)

Half life ,t 12

when x = a2

kt 12

=a2

a(a− a2 )

⇒ t 12

= 1ka

Example

In a certain thermolecular reaction the decrease x in concentration of asubstance R is given by dx

dt = k(a− x)3,where k is the reaction rateconstant and a is the initial concentration of R. Find the concentration attime t.



Solutiondxdt = k(a− x)3∫

dx(a−x)3 = k

∫dt

12(a−x)2 = kt + C

At t=0, x=0 ⇒ 12(a−0)2 = k(0) + C ⇒ C = 1

2a2

Thus 12(a−x)2 = kt + 1

2a2

⇒ a2

(a−x)2 = 1 + 2a2kt

⇒ (a− x)2 = a2

1+2a2kt ⇒ a− x =√

a2

1+2a2kt

⇒The concentration of the substance R is given as(a− x) = a√

1+2a2kt



RADIOACTIVE DECAY

When radioactive substances decay, the number of atoms that decay in afixed time period is proportional to the number of atoms at the start ofthat period. Thus the rate of decay of a radioactive substance isproportional to the number of atoms N present at time t. If the constantof proportionality is λ the decay constant and initially are NO atomspresent, thendNdt ∝ N

⇒ dNdt = −λN∫

dNN = −λ

∫dt

lnN = −λt + C

When t=0, N =No ⇒ C = lnNo



⇒ lnN = −λt + lnNo

lnN − lnNo = −λtln N

No= −λt

NNo

= e−λt

N = Noe−λt

Example 1

Carbon 14 , one of the three isotopes of carbon is radioactive and decaysat a rate which is proportional to the amount present. Its half life i 5570years. If 10 grams were present originally, how much will be left after2000 years.



SolutiondNdt ∝ N

⇒ dNdt = −λN∫

dNN = −λ

∫dt

lnN = −λt + C

When t=0, N =No ⇒ C = lnNo

⇒ lnN = −λt + lnNo

lnN − lnNo = −λtln N

No= −λt

NNo

= e−λt

N = Noe−λt



At half life, N = No

2 , t 12

= 5570

⇒ No

2 = Noe−λ(5570)

12 = e−λ(5570)

−5570λ = ln 12

λ = − 15570 ln 1

2 = 0.000124

⇒ N = Noe−0.000124t

At t=2000 years ,No = 10grams

N = 10e−0.000124(2000)

= 7.8036 grams ' 7.8 grams.



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