Mathematics mathematics three 3 Lec 2.pdf

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1. Chapter (1)
Introduction to Differential Equations

1.1 Differential Equation (DE) :
Definition (1.1) :

A is an equation containing the derivatives
(or differentials) of one or more dependent variables with respect

to one or more independent variables .

* For examples ,

2. az à |
ds 9
d'u Yu d'u.

dx? ay? a? u

22

Notes :

(i) If DE contains ordinary derivatives , it is called ordinary
differential equation (ODE). Equations (1) and (2) are
ordinary differential equations .

(ii) If DE contains partial derivatives , it is called partial
differential equation (PDE). Equations (3) and (4) are
partial differential equations .

3. 1.2 Order of Differential Equation :
Definition (1.2) :

The of differential equation is the order of the highest
derivative present in the differential equation .

* Equation (1) is of second order.
is of fourth order.

* Equation

(
* Equation (2

(

(

)
)

3) is of first order.
)

* Equation (4) is of second order.

4. 1.3 degree of Differential Equation :
Definition (1.3) :
The of differential equation is the greatest power of the

highest derivative present in the differential equation after
writing it in the form of power series (standard form) .

* Equation (1) is of first degree.

* Equation (2) is of third degree .

(2)
(2

* Equation (3) is of first degree.
(

* Equation (4) is of first degree .

5. 1.4 Linear Ordinary Differential Equation :
Definition (1.4) :

An nth-order

a, (x)

d'y dry

+a,_,(x +...
dx" " À a (1)

dy
+4, a +a) (x)y =F (x)

where a, #0 is said to be 90,

(i) The dependent variable (7) and all of its derivatives

(s 1 ㅠ 뻐 } are of the first degree.

(ii) No product of the dependent variable (y ) and any of its

derivatives 6 Ty ayy”) } is present in the equation .

6. 6
If one of the three previous conditions is not satisfied, then the ordinary

7. Example (1):

Determine whether each of the following equations are linear or
nonlinear , name the dependent and independent variables , and
state the order and the degree of each equation .

d

2

y
a +5 +6y=0.
(a) a ae tee

dy Y
(b) ES +sin y =x?.
dx

4
d’y d’y dy
€, +2 + =0.
m (E | dx? Y dx

8. Solution

Dependent |Independent| Order Degree | Linear or
variable variable nonlinear

9. EXERCISE

Determine whether each of the following equations are linear or
nonlinear , name the dependent and independent variables , and
state the order and the degree of each equation .

2 dy dy

“ar di
dy dy dy dy dy
dé dd à? 200
Py

a E By
5. => +sin(t + y) =sint =
de 7

dy dy
dt? dt

dy 5 x
5 > +t— + (cos*t)y=t
dé dt 7

10. 1.5 Solution of Differential Equation :
Definition (1.5) :

of differential equation is an equation containing the
dependent variable ( for example y ) and the independent
variable ( for example x ) and free from derivatives , which

satisfies the differential equation .

11. Notes :

(i) The general solution of the differential equation is an
equation consists of a family of solutions and containing one
or more arbitrary constants are equal to order of differential
equation .

(ii) The particular solution of the differential equation is any
solution free of arbitrary constants .

(iii) y =F (x) is an explicit solution of ODE.

(iv) G(x,y)=0 is an implicit solution of ODE.

12. Example (2):

Show that y =x? - 2 is an explicit solution of the differential
x

2
equation アリ ーー テア =0.
x

Solution

Remember that :

13. 2
* Since Lift side = y "-S y
x

2 20 , 3
=|2 x
( 3) 2 BE
2 2
oe
x x

=0 = Right side

14. Example (3):

Show that y?-x’+8=0 is an implicit solution of the
2

3
differential equation y' = > on the interval (2,00) .
y

Solution

* When we solve y? —x* +8 =0 for y we obtain

yt vx? -8
which is defined on | 2,00)
Remember that :
„If (x)= Vx square root function

Domain: x 20 or [0,0)

15. * Since

which is defined on (2,0)

* Since Lift hand side =y'

= Right hand side

16. * Then y?’-x’+8=0 is an implicit solution (general

3 了 X2
solution) of the differential equation y' = y on the interval

y
(2,0).

EXERCISES:

4:

Verify that y = xt is a solution of the differential equation

dy Sa
„xy: in (一 co, oo)

© [Verify that y = xe* is a solution of the differential equation
y"-2y'+y=0 in(-0,00)

17. 1.6 Initial-Value and Boundary-Value Problems :

Definition (1.6) :

A differential equation along with subsidiary conditions on the
unknown function ( for example y ) and its derivatives , all given
at the same value of the independent ( for example x ) ,
constitutes an initial-value problem (IVP) . The subsidiary
conditions are initial conditions (1.C.) .

18. Definition (1.7) :

If the subsidiary conditions are given at more than one value of
the independent value ( for example x ) , the problem is a
boundary value problem (BVP) , and the conditions are
boundary conditions (B.C.) .

19. Example (4):
Show that y =sinx-cosx is a solution of the Initial-value
problem (IVP)
y"+y=0 , y(0)=-1 , y'(0)=1.
Solution

* Since

y =sin x —cos x

y' =cosx +sinx

J

" =—sinx +cos x

* Since Lift hand side = y "+ y

= (- sin x + cos x)+(sinx — cos x)

=0 = Right hand side

20. * Then y =sinx-cosx is a solution (general solution) of
differential equation y"+y =0. (1)

* Since y (0)=sin0 -cos0 =0 -7 = ㅡ 7 , then the first initial

condition (I.C.) is satisfied . (2 )
Remember that:
|sin0 =0 , cos0=1

* Since y'(0)=cos0+sin0 =1+1=1 , then the second
initial condition (1.C.) is satisfied . (3)

* From (1), (2) and (3) then y =sinx—cos x is a solution
(general solution) of the given IVP .

21. Chapter (2)
First-Order Ordinary Differential Equations

Definition (2.1) :

The general form of ordinary differential equation of the first
order is

v(x,y,y’)=0

= er)

dx

22. 2.1 Separable Equations :
Definition (2.2) :

A first order differential equation of the form
dy
2x = IDA)
where g(x)h(y) are functions of x &y only, respectively. is called separable equation.
Rearranging this equation, obtain

[roy J owe > [pay = | guar

x
Solve 2 ==
dx y

on By separable variable
y.dy=x.dx => Jo = [re >
Then y = +yx? +A where A is constant

23. y.

d
2: Solve v=
dx

By separable vais
dy _ dx dy dx
7 > $3 E > Iny =Inx+Inc

1 ] 1 In==1
ny —Inx =Inc nz nc

Then y=cx

3: Solve the initial-value problem Fr
n By separable variable
y.dy=-x.dx > fre [re >

y? +x? =2c butx=4&y=3 then9+16=2c >c=12.5
y?+x?=25

24. Example 4: Solve 2 = cos 5x
Solution By separable variable
dy =cos5xdx => | = | cos 5x ax

sin 5x

Then yas

Example 5: Solve (1+x)dy — ydx = 0
Solution (1 + x)dy — ydx = 0 (1+ x)dy = ydx

dy d d d
TE DE [?- =
y

y 1+x Loa
> Iny=In(it+x)+c > y=(1+x)e
Then y=(1+x)c,

25. Example 6: Solve x — y = 2x?y
Solution : ie =yt2xy = x = y(1 + 2x2)
>. (1+ 2 à dy

y

1
= に + 2x) dx
x

d d 1
ja [€ 5 > [I = [24 +2 f sas

x
> Iny=Inx+ 23 +0 > Iny=Inx+ x?+c

Then y= 4xe ‚where A = constant = 60

26. Example 7: Solve
y?

a

Solution :

y?-4

=a > [565% dl

Voz»

1 1 1
ィ ツー2) 一 Im ツ +2) ニャ +c > mí

y-
> (ーッ

=> y-2=(y+2)ce*

=>

) = 4x + 4c >>

| if ay= [a
E Jo
y=2
y+2

=e (4x+4c)

arte
アー

yt2
where ci = constant = e*°

_2(1+c,e*)

中 Ge

27. 27
1) 2)
3) 4)
5) 6)
Solve the given differential equation by separation of variables.

28. Equations Reducible to Separable :

Let y = f(az + by +c) £
From (1) and (2), we have
dz

ae Oo Fe )> E =p )+a
서게

by separable variable, we have

anu solution is
dz

ae
Hoan Nr

dy
dz

Example 1: Solve the differential equation (x + y)?

ody

Solution: The given DE is (x + y) 1
dz

=4,

dy 4

>=; の
de (x+y)?

29. Lett=x+y
dy
2 ~
dy _ dt
de dx
dt 4

=1
dx 2

P+4-4
> 一 一 一 一 一 |
J. t+

1

+ 1

P+?
æt-2tan (ff) =x+c

Ur+y)=c
2

= y — 2tan

30. 30
2.2 Linear Equations :
Definition (2.3) :
Procedure of Solution:

31. d
Example 1: Solve ae

Solution: 2 — 3y =6 p(x)=-3 & f(x) =6
integrating function = I(x) = e7/3dx = e 3x
The general solution

—6
20 = i 6.e%*dx+c > ye*= Er +c

-3x — ー フ っ -3 = =
> 76 *= le "He > y= 24 ar

> y=-2+ce?*

32. Example 2: Solve

Solution: > p(x) = - & f(x) = =

x

2
= eszdx = e2Inx _ 22

integrating factor = I(x)
The general solution
라 으 | 2 2 에 |
yx? =] —x"dx+c => y.x*=2] xdx+c
* 2 2
> yee = a= +c > y=

=> yalter”

33. Example 3: Solve x + (3x + 1)y =e"

gas ~: (SRE) war dy ( +2) dde
Solution : nt x DM x ur 34 72?
1

= p(x) =3 ㅜ ㅜ & fe) = Le"

1
; y 3+2a
integrating factor = I(x) = ell +) X 一 e3x+inx 一 xe3x
The general solution

1
Xe" ダー fre xeax + c + pets fa: +c

と
> yxeX =x+c pe “26

34. Example 4: Solve x{2—4y = x%e*
dx

R d a
Solution : =$ >2- (5) y=x%e*

x dx x,

> 700 = 一 & f(x) =x5e*

integratin, r= = sí à) = etlnx = xt
grating facto: I(x) e
he general solution

ya * = faserx-tax + o => ya? = fre dx +e

=> yx t=xe*-e* + 0 > y=x5ex —xte*+cx4

35. Example 5: Solve (x? -92 +xy=0

al 29 __* _
Solution: + sy = 0 p(x) Ed e f(x)=0

x Lt
integrating factor = I(x) = ejer = 0206 9 = VX2 一 9
The general solution

y. 8-5 = fo, x?-9dx+c > y [79 = [o.ax +6

> wWx-9=c >

vx2-9

36. Example 6: Solve the initial-value problem 은 +2xy =x, y(0)=1

Solution : =. +2xy=x の (%) =2x € fox

ott
integrating factor = I(x) = el 2xdx = eT = ex?
The general solution

1 2
ye? = [xe¥dx+c > pe“ => e* +e

1
ッ ニ ュ オ ce で Put x=0 € y=1

1 2
> ニラ (1+ er)

37. 37
Exercises:
Find the general solution of the following differential equations: