First-order linear equations
description
Transcript of First-order linear equations
First-order linear equations A first-order linear equation has the general form
If the equation is called homogeneous; otherwise
it is called inhomogeneous.
For example, is a linear equation, and an
inhomogeneous one, since it can be written as 2.y
yx
( ) ( )y P x y Q x
2xy y x
( ) 0,Q x
Integrating factor method To solve the first-order linear equation
we multiply the equation by a suitable function I(x):
If the factor I(x) is chosen such that
then equation (2) becomes
which can be solved by
( ) ( ) (1)y P x y Q x
( )( ( ) ) ( ) ( ) (2)I x y P x y I x Q x
( )( ( ) ) ( ( ) ) (3)I x y P x y I x y ( ( ) ) ( ) ( ),I x y I x Q x
1( ) ( ) ( ) ( ) ( ) .
( )I x y I x Q x dx C y I x Q x dx C
I x
Integrating factor method Thus the key point to solve equation (1) is to find I(x) such
that equation (3) holds true:
This is equivalent to
which is a separable equation for I(x). Its solution is
Simply taking C=1, we call an integrating
factor of equation (1).
( )( ( ) ) ( ( ) ) ( ) ( ) .I x y P x y I x y I x y I x y ( ) ( ) ( ) (4)I x P x I x
( )( ) .
P x dxI x Ce
( )( )
P x dxI x e
Example Ex. Solve the equation Sol. An integrating factor is
Multiplying I(x) to the equation, we get
Ex. Solve
Sol.
2 23 6 .y x y x 2 33
( ) .x dx xI x e e
3 3 3 32 2 2( 3 ) 6 ( ) 6x x x xe y x y x e e y x e 3 3 3 326 2 2 .x x x xe y x e dx e C y Ce
sin.
y xy
x x
1
( )dxxI x e x
coscos .
C xxy x C y
x
sin ( ) sinxy y x xy x
Example Ex. Solve the equation
Sol. Not a linear equation? What if we treat x as dependent
variable and y as independent variable:
2.
2
dy y
dx x y
22 2.
dx x y xy
dy y y
22( )
dyyI y e y
2 3 12
dxy xy ydy
2 1 2( ) ln | |dy x y y x y C
dy
2( ln | |) .x C y y
Example Ex. Solve the equation
Sol.
Ex. Solve the initial value problem
Sol.
2cos tan 0.dyt y tdt
2sec tan( )tdt tI t e e
tan tan 1.ty Ce t
, (2) 1.2 ln
yy y
y y y x
1
( )dyyI y e y
2
ln .x y yy
Example Ex. Solve the initial value problem
Sol.
2 | |, (1) 1.y y x y
21
1 10 ;
2 4xx y C e x
22
1 10
2 4xx y C e x
21
3(1) 1
4y C e
22 1 2
1 1 3 1(0 ) (0 )
4 4 4 2y y C C C e
2 2
2 2
3 1 1( 0)
4 2 4 .3 1 1 1
( ) ( 0)4 2 2 4
x
x
e x xy
e e x x
Homework 22 Section 9.3: 7, 10, 15
Section 9.6: 12, 14, 19
Page 648: 1