2x od i㱺 x run 2 -5=1 5=1 · 2019. 11. 19. · chi differential equation: damnboyars n hi 2x-5=1...
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CHI Differential Equation : damnBoyarsN hi 2x - 5=1 ⇒an x ainnhirwmrnfuos
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2x = 6
X ofbpinnovorcsosrw 2×-5=1
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bad y = f@xk4x)dx = 2×3 - 2x't C
rawinsonde : orsrnrndstowhvakingo worms
ovoiuprpmvrnnghsrwmosoynrot.IN oisoivvasrmsisoyovoFE day, = 2X irwmrgouirrohsiuuus
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(First- order differential equation )
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abrasives
( second - order differential equation )
d¥×+y(d¥f②=c0sx : rwrwirsoyouroisoiuuw ,
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omoifsoy.hr Gnats (ordinary differential equations " ODES(⇒ rainbow
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wud eiov (patrol differential equation : PDEs ) )
9Differential Equations
In this chapter, we introduce two methods for solving some form of the first order of differential
equations (ODEs). First, we introduce some basic definitions of ODEs. We, then, solve the particular
ODEs in the forms of Separable equations and Linear first order ODEs . Lastly, some examples
of linear first order ODEs.
9.1 Introduction to Ordinary Differential Equations
Consider the equation, y = 2x3 − 2x2 + 5. By differentiation, it can be shown that
dy
dx= 6x2 − 4x. (9.1)
Similarly, for a function p(x) = 10000e−0.04x, we have
p′(x) = −400e−0.04x. (9.2)
These equations are example of differential equations .
In general, an equation is a differential equation if it involves an unknown function and one or
more of its derivatives. Other examples of differential equations are
dy
dx= ky, y′′ − xy′ + x2 = 5,
dy
dx= 2xy
The first and third equations are called first-order equations because each involves a first deriva-
tive but no higher derivative. The second equation is called a second-order equation because it
involves a second derivative and no higher derivatives. In general, the order of a differential equation
is the order of the highest derivative that it contains.
169
170
9.2 General and Particular Solutions
A solution of differential equation is the function which matches the differential equation.
Example 9.1 Show that the function y = ex is a solution of
dy
dx− y = 0
Example 9.2 Show that, for any constant C, the function y = ex − x+ C is a solution of
dy
dx= ex − 1
Remark:
• The general solution of a differential equation is a solution that contains all possible solutions.
The general solution always contains an arbitrary constant.
• The particular solution of a differential equation is a solution that satisfies the initial condi-
tion of the equation. A first-order initial value problem is a first-order differential equation
y′ = f(x, y) whose solution must satisfy an initial condition y(x0) = y0.
Academic year 2019 206111: Calculus 1
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LHS.RHS .
4- et
da =e×ELIE dx
Doom DI- y = ex -et = 0=42 . His
.DX
: . y--et bournMarra runs day, -4=0
nLHS RH.
S.
An y-
- ex-Xtc
L.His .DI
=ex - I = R .
His.
DX
:. ya ex
-X + C iahroinmowa dIa×= ex - I #
171
Example 9.3 Find the particular solution of
dy
dx= ex − 1, y(0) = 1.
Example 9.4 Show that the function
y = (x+ 1)− 1
3ex
is a solution to the first order initial-value problem
dy
dx= y − x, y(0) = 2/3.
206111: Calculus 1 Academic year 2019
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initial condition
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an Ex 9.2 , general solution as day, = ex - I pdo/y=e×-xtCy
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-
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. y-
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et- X tC 186 Armourown two y 107=1
{① y Iwamuramountain
② y NoraJo, n'Ul5ow7vn45NnV7ha
L. His f.Hrs.
-
① y -- Cxtn) -Iet ② y sonantsN y lo) = 2/3 ?
l day = 1 - Iz ex y Cx) = txt ) - { ex
y ( o) = OH - tze°
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.= a- tze't
I. L.H .
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172
9.3 Separable Equations
We will now consider a method of solution that can often be applied to first-order equations that are
expressible in the form
h(y)dy
dx= g(x). (9.3)
Such first-order equations are said to be separable . The name separable arises from the fact that
(9.3) can be rewritten in the differential form
h(y)dy = g(x)dx (9.4)
in which the expressions involving x and y appear on opposite sides. To motivate a method for solving
separable equations, assume that h(y) and g(x) are continuous functions of their respective variables,
and let H(y) and G(x) denote antiderivatives of h(y) and g(x), respectively. Consider the results if
we integrate both sides of (9.4), the left side with respect to y and the right side with respect to x.
We then have
∫h(y)dy =
∫g(x)dx, (9.5)
or, equivalently,
H(y) = G(x) + C (9.6)
where C denotes a constant. We claim that a differentiable function y = y(x) is a solution to (9.3) if
and only if y satisfies (9.6) for some choice of the constant C.
Academic year 2019 206111: Calculus 1
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173
Example 9.5 Write these first-order differential equation in the separable form.
Equation Form h(y) g(x)
dy
dx=
x
y
dy
dx= x2y3
dy
dx= y
dy
dx= y − y
x
Example 9.6 Find the general solution of
dy
dx=
x
y.
Example 9.7 Find the general solution of
dy
dx= yex.
206111: Calculus 1 Academic year 2019
hey ) day, = gcx )
CO 4daL -- x y X
←O # daff -- X'
at, XZ
III -- I f I
y1dIa×=i - I Iy I - Ixta:* .si -⇒
ionium differentiae : yay =×d×
→Y¥oisninsmnsooMs : fydy = fxdx
#Ia -
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=. yay
=exa×→T¥{ Iydy =fe×dx von cairn
✓
butyl = exy.ae#i*-oy=ee4i=ceeQB-
174
Example 9.8 Find the general solution of
dy
dx=
√xy .
Example 9.9 Find the general solution of
dy
dx=
xy + y
xy − x
Example 9.10 Solve the initial value problem
dy
dx= −4xy2, y(0) = 1.
Academic year 2019 206111: Calculus 1
trydy -_r×d×→Fyd¥
ftfydy - frxdx
#
⇒ yE=x÷Ic ⇒ y=fz¥ '
→ a¥=¥Yiday --¥11 )
( dy =(x)dxIfi - f)dy = fat DX
He
-
↳ y1zdIa×= - 4X
fyIzdF× = f-4xdX
-19=-2×2-2-1 -E4107 - I ', -1
,= -2105 -1C =D C= - I
⑦ ; - ly = -2×2 - 1
ty = 2×2+1
i.y=f×
175
Example 9.11 Solve the initial value problem
yy′ − (x2 + 1) = 0, y(4) = 2.
Example 9.12 Solve the initial value problem
(4y − cos y)dy
dx− 3x2 = 0, y(0) = 0.
206111: Calculus 1 Academic year 2019
4daL = XII
ydy =Cx4Ddx
fydy =
fcxhhdxykh.ee;
¥=x'If, = t¥+4tc ⇒ 2-4-634 =C
c = 6--123-64=-73i.t=Itx-EC #
(44 - cosy )day, = 3×2
(44 - cosy )dy =3x2dX
((4y - cosyldy =f3x2dX2y2_siny=x3T
4101=0 , O - Sino = Otc ⇒ C -- O
kimono : 2y2-siny- #
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y′ − 2x(1 + y2) = 0, y(2) = 1
L. H. S. RHS .
--
y =xe×
y' =Xe×te×
y"= Xe× text ex =Xe×t2e×
: . LHS : y"- y =fXe×t2e×)- xe×=2e×
f. His = E.
.
. t.H.s.FR . H. S.⇒ y=xe×7d nhspinnovvbrsrmr
bdyd-×- 2×(1+42)=0
day, = 2XCity )
+÷z dy = 2X dx
1¥12 dy = fzxdx
arctany = x't C
y (21=1 ; arctan I = 4 t C
IT
-4= 4 + C
C = I 44-
i. particular solution : arctary-x-I.TL! #