2x od i㱺 x run 2 -5=1 5=1 · 2019. 11. 19. · chi differential equation: damnboyars n hi 2x-5=1...

CHI Differential Equation : damnBoyarsN hi 2x - 5=1 ⇒an x ainnhirwmrnfuos

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2x = 6

X ofbpinnovorcsosrw 2×-5=1

run-- 2137-5 = 6 -5=1 gumbosod i⇒ . www..mn.. .

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(un y ri diff linin bxh-4xmdidwdon.ru

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 )

y' '

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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

Armours raw DES 82kW

(1) general solution caiman'Du )

( n'amountMv: sinorounsuwnnnhidlor )(z ) particular solution cpirmouiom - I

(a'nmoundinnow metrics'ovN winewa )

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

initial-valaepwblan-ianoinmouaton.nuwaioowlvpihowoiuanti

initial condition

-

an Ex 9.2 , general solution as day, = ex - I pdo/y=e×-xtCy

'

-

winnow £ : 4101=1 ; I = ef- Otc

.

'

. y-

-

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°

r y - x-- hair 'sex -X ) = , - z = } qq.ygiomy.ua,...

.= a- tze't

I. L.H .

S = R.tl. S.

.

'

, y = 4th ) - Tze' ieboinmowcs IVI

.#

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.


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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.


dy

dx= yex.


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 -

- II + few. civil?

=. yay

=exa×→T¥{ Iydy =fe×dx von cairn

✓

butyl = exy.ae#i*-oy=ee4i=ceeQB-

174


dy

dx=

√xy .


dy

dx=

xy + y

xy − x

Example 9.10 Solve the initial value problem

dy

dx= −4xy2, y(0) = 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


yy′ − (x2 + 1) = 0, y(4) = 2.


(4y − cos y)dy

dx− 3x2 = 0, y(0) = 0.


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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�� ¦ª�°�ªnµ ¢́��r��́ y = xex Á�È��¨Á�¨¥�°�¤�µ¦Á�·�°�»¡�́�r y′′ − y = ex ®¦º°Å¤n

�� ®µ�¨Á�¨¥�°��́�®µ�nµÁ¦·É¤�o�

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! #

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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