Differential Equations Definitions

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Differential Equations 2301312 ISE 1

Differential Equations

{

Introduction

{First Order Differential Equations

{ A differential equation is an equation that

contains an unknown function, y(x), and some of its derivatives, dy/dx or y’(x).

{ The order of a differential equation is the order of the highest derivative that occurs in the equation.

{ Examples:

1 3

8 (3)

) 1 ( (2)

1 2

(1)

3 4

2 2

2 3

3

−

=

⎟ +

⎠

⎜ ⎞

⎝

− ⎛

⎟⎟⎠

⎜⎜ ⎞

⎝

⎛

=

− +

−

+

=

x dx xy

x dy dx

y d

e y dt t

t dy dt

y d

y dx x

dy

t

Definitions

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Derivatives

{ Let y be a function of x. The

derivative of y with respect to x, at the point a is

{ The derivative y’(t) is the rate of change of y w.r.t. t. If y(t) is the distance function at time t, then y’(t) is the velocity at time t.

. ) lim

( ) lim (

: ) ( )

(

' 0 0 x

y x

a y x a a y

dx a dy

y x x Δ

= Δ Δ

− Δ

= +

= Δ → Δ →

Solution

{ A function f is called a solution of a differential equation if the equation is

satisfied when y = f(x) and its derivatives are substituted into the equation.

{ If f is a solution of y’ = xy, we must have f’(x) = xf(x)

for all values x in some interval.

{ To solve a differential equation, we are expected to find all possible solutions of the equation. For example,

x C y = +

2

x

y ' =

has the solutions

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Example

{ Show that every member of the family of functions

is a solution of the differential equation

y’ = (y² -1)/2.

{ Find a solution of the differential equation

y’ = (y² -1)/2 that satisfies the initial condition y(0) = 0.

( ) 1 1

t t

y t ce

ce

= +

−

(4)

General solution

{ Consider

then

y = e

^2xis a solution of this differential equation.

{ In addition,

y = 2e

^x

+ e

^2xis also one of the solutions.

{ We say that

y = Ce

^x

+ e

^2x is a general

solution of the differential equation, where C is any real number.

{

y = 2e

^x

+ e

^2xis a specific solution for the differential equation.

e x

dx y

dy ₂

=

−

Ordinary and Partial Differential Equations

{ Ordinary Differential Equations (ODE)

(The unknown function depends on a single independent variable.)

z y’ = (y² -1)/2 as y=y(t).

z y’ = xy as y=y(x).

{ Partial Differential Equations (PDE)

(The unknowns function depends on several independent variables.)

Equation) (Wave

) , ( )

, (

Equation) (Heat

) , ( )

, (

2 2 2

2 2

2 2 2

t t x u x

t x a u

t t x u x

t x a u

∂

= ∂

∂

= ∂

∂

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First Order Differential Equations

{ Separable Equations

z y’=f(x)g(y) or in differential form

z M(y)dy = N(x)dx

{ Linear First Order Equations

{ y’+P(x)y=Q(x)

z Bernoulli equation

{ y’+P(x)y=Q(x)yⁿ , when n ≠ 0,1

Separable equations

{ A separable equation is a first-order

differential equation in which the expression y’ can be factored as a function of x times a function of y. Namely,

or equivalently in differential form

{ So that all y are on one side of the equation and all x are on the other side. Integrate both sides,

), ( ) (x f y dx g

dy =

. ) ( )

(y dy N x dx

M =

. ) ( )

∫

^h(^y ^dy ⁼

∫

^g ^x ^dx

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Example

z Solve the differential equation

z Find the solution of the above equation that satisfies the initial condition y(0) = 1.

z Solve the differential equation

z Solve the equation y’ = xy.

2 . y

x dx dy =

sin . 3 ³

y y

x dx

dy

= −

Exercises 1.

{ Solve the given differential equations.

{ Solve the given initial value problems.

2 2 2 2

/ 1 2

3 2

2

. 1 6 .

5

) 2 3 /(

) 1 3

( ' . 4 )

1 ( ' . 3

) 1

( / '

. 2 /

' . 1

y x dx

dy e

y e x dx dy

y x

y y

xy

x y

x y y

x y

y x

= + +

= −

+

−

=

−

=

+

=

−

0 ) 0 ( ),

2 3 /(

) 2

( ' . 3

1 ) 0 ( ,

2 ' . 2

1 ) 0 ( ),

6 3

/(

) 3 1 ( ' . 1

2 2

= +

−

=

= +

=

− +

=

y y

e y

y xy

y y

y x

y

x

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A first order linear differential equation

{ A first order linear differential

equation is one that can be put into the form

where P and Q are continuous functions on a given interval.

{ For example, xy’ + y = 2x is a first order linear differential equation but not

separable.

{ Since (xy)’=xy’ + y, we have xy = x² + C.

) ( )

( x y Q x dx P

dy + =

Integrating factor

{ To solve the first order linear differential equation

it is suggested to multiply a suitable I(x) called an integrating factor, so that

Hence,

( ) ( ), dy P x y Q x

dx + =

( ( ) )' ( ) dy ( ) ( ) ( ).

I x y I x P x y I x Q x dx

⎛ ⎞

= ⎜ + ⎟ =

⎝ ⎠

C dx x Q x I y

x

I⁽ ⁾ =

∫

⁽ ⁾ ⁽ ⁾ +

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Integrating factor

{ The property of I is

{ This is a separable differential equation,

{ Hence, the integrating factor is

and we can set constant A=1.

).

( ' ) ( )

( x P x I x

I =

, )

(_x ₌ _Ae

∫

^P⁽^x⁾^dx

I

∫

=

. ) ( ln

, ) 1 (

dx x P I

dx x P I dI

Summarize

{ To solve the linear differential equation y’+ P(x)y = Q(x),

multiply both sides by the integrating factor and integrate both sides.

{ The general solution is

= e ∫

^P ^x ^dx

x

I ( )

⁽ ⁾

(

⁽ ⁾ ⁽ ⁾

)

^.

) ( ) 1

( I x Q x dx C

x x I

y =

∫

+

Basic integral formula

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Basic Integral Formula

∫

+ + =

•

+

= +

−

=

•

−

=

•

≠ +

=

•

+

=

•

−

≠ + +

=

•

− +

C x x dx

C x dx

x C

x dx

x

du v uv

dv u

a a C

dx e e

C x x dx

n n C

dx x x

ax ax

n n

) ( 1 tan

1

sin cos

, cos

sin

) parts by n integratio (

, 0 ,

,

|

| 1 ln

, 1 1 ,

1 2

1

Example

{ Solve the differential equation

{ Find the solution of the initial-value problem

{ Solve y’+ 2xy = 1.

. 6 3x²y x² dx

dy + =

. 2 ) 1 ( and 0

, 1

2y'+xy = x > y =

x

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Exercises 2

{ Find the solution of the given initial value problems.

0 ,

1 ) 2 (ln ,

) 1 ( ' . 6

2 ) 0 ( ,

2 ' . 5

0 ,

0 ) 1 ( ,

4 ' .

4

0 ,

) 1 ( ,

1 2

' . 3

0 ) 1 ( ,

2 ' . 2

1 ) 0 ( ,

2 '

. 1

2 2 3

2 2 1

2 2

>

=

= +

+

=

−

<

=

−

= +

>

= +

−

= +

=

= +

=

−

t y

t y t

ty

y e

y y

t y

e y t y t

t y

t t t ty

y te

y y

y te

y y

t t t t

Bernoulli equation

{ A Bernoulli differential equation is of the form

If n = 0 or n = 1, the Bernoulli differential equation is just linear.

{ We transform this differential equation to a linear equation by setting u = y^1-n. We get

. ) ( )

(x y Q x yⁿ dx P

dy + =

).

( ) 1 ( )

( ) 1

( n P x u n Q x

dx

du + − = −

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Example

{ Solve the differential equations

z

z t²y’+2ty –y³ = 0, for t > 0.

' y

3

x y + y =

2

2 xy x3y

dx

dy + =

Answers

{ Exercises 1:

{ Exercises 2: