Download first-order linear equation

First-order linear equations
A first-order linear equation has the general form
y  P( x) y  Q( x)
If Q( x)  0, the equation is called homogeneous; otherwise
it is called inhomogeneous.

For example, xy  y  2 x is a linear equation, and an
y

y

 2.
inhomogeneous one, since it can be written as
x

Integrating factor method
To solve the first-order linear equation
y  P( x) y  Q( x) (1)
we multiply the equation by a suitable function I(x):

I ( x)( y  P( x) y)  I ( x)Q( x) (2)
If the factor I(x) is chosen such that
I ( x)( y  P( x) y)  ( I ( x) y) (3)
then equation (2) becomes ( I ( x) y )  I ( x)Q( x),
which can be solved by
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:
I ( x)( y  P( x) y )  ( I ( x) y )  I ( x) y  I ( x) y.
This is equivalent to I ( x) P( x)  I ( x) (4)
which is a separable equation for I(x). Its solution is
P ( x ) dx

I ( x)  Ce
.
P ( x ) dx

 Simply taking C=1, we call I ( x)  e
an integrating

factor of equation (1).
Example
Ex. Solve the equation y  3x 2 y  6 x 2 .
2
3
x
dx
x3

 Sol. An integrating factor is I ( x)  e
e .
Multiplying I(x) to the equation, we get
x3
2
2 x3
x3
2 x3
e ( y  3x y)  6 x e  (e y)  6x e

 e y   6 x e dx  2e  C  y  2  Ce
x3


2 x3
Ex. Solve y 

x3
 x3
.
y sin x

.
x
x
1
dx
x
 x  xy  y  sin x  ( xy )  sin x
C  cos x
 xy   cos x  C  y 
.
x
Sol. I ( x)  e
Example

dy
y

.
Ex. Solve the equation
2
dx 2 x  y
Sol. Not a linear equation? What if we treat x as dependent
variable and y as independent variable:
dx 2 x  y 2 2 x


 y.
dy
y
y

2
  y dy
dx
I ( y)  e
y y
 2 xy 3   y 1
dy
d

( y 2 x)   y 1  y 2 x   ln | y | C
dy
 x  (C  ln | y |) y 2 .
2
2
Example

dy
Ex. Solve the equation cos t
 y  tan t  0.
dt

Sol. I (t )  e
2
 sec2 tdt
 e tan t
y  Cetan t  tan t  1.


Ex. Solve the initial value problem
y
y 
, y (2)  1.
2 y ln y  y  x
Sol. I ( y )  e 
1
dy
y
y
2
 x  y ln y  .
y
Example


Ex. Solve the initial value problem y  2 y | x |, y (1)  1.
1
1
3 2
Sol. x  0  y  C1e  x  ; y (1)  1  C1  e
2
4
4
1
1
2 x
x  0  y  C2 e  x 
2
4
1
1
3 2 1
y (0)  y (0)  C2   C1   C2  e 
4
4
4
2
1
 3 22 x 1
e
 x
( x  0)

4
2
4
 y
.
( 3 e2  1 )e 2 x  1 x  1 ( x  0)

 4
2
2
4
2 x
Homework 22

Section 9.3: 7, 10, 15

Section 9.6: 12, 14, 19

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