Sec 1.5: Linear First-Order DE

1

Sec 1.5: Linear First-Order DE

Definition 2.2

:Example

)()(')( 01 xgyxayxa

1

A 1st order De of the form

Sec 2.3

is said to be linear first-order equation.

03' yy 2 363' yy xexyxy 64'

)()(' xQyxPy

2

How to Solve ?

Q(x)P(x)yy '

Sec 1.5

Method of Solution:

Step 1Rewrite into standard form

(coeff of y’ is 1).

dxxP

ex)(

)(Step 2Find the Integrating Factor =

( ignore constant of integration)

----- (1)

Step 3 Multiply (1) by ( check: )][ LHS yDy](x)[dx

dx

Step 4Integrate both sides:

)()( xQxy](x)[dx

d ( DONOT forget constant of integration)

:Example

1 03' yy 2 363' yy xexyxy 64'

x

x

3

Remark: linear in y or linear in x

:Example

Sec 2.3

1)0(8

11' 3/

y

eyy x

xxyyx 63')1( 2

Solve the IVP

:Example Find a general solution of

4

:Example Find a general solution of

2

1

yxdx

dy

Q(x)P(x)yy '

)()( )(

1)( CdxxxQ

xxy

The solution of the above DE is given:

dxxP

ex)(

)(where

Derivation

5

00 )(

'

yxy

Q(x)P(x)yy

)()( )(

1)( CdxxxQ

xxy

If the function and are continuous on the open interval I containing the point , then the initial value problem

dxxP

ex)(

)(where

Theorem 1:

Theorem 1: The Linear First-Order Equation

has a unique solution y(x) on I, given by the formula in Eq(6) with an appropriate value of C.

xP xQ

0x

(6)

Remarks:

1. Theorem 1 gives a solution on the entire interval I

2. Theorem 1 tells us every solution is included in the formula (6)

3. Theorem 1 tells us that the general solution is given in(6)

4. Theorem 1 tells us that a linear first-order DE has no singular sol

6

:Example

Sec 2.3

10

2

)1(

sin'

yy

xxyyx

Solve the IVP

7

Derivative and Integration

D[x^3+3 x,{x,2}]

x][xdx

d33

2

2

Integrate[ x^2 , x]

dxx2

syms x

diff(x^3+3 x)

MA

TH

EM

AT

ICA

syms x

int(x^2,x)

MA

TL

AB

8

9