LECTURE 7: INTEGRATING FACTOR AND EULER’S METHOD
MINGFENG ZHAO
September 17, 2014
To solve an exact equation M (x, y) + N (x, y) y 0 = 0:
1) Compute My (x, y) and Nx (x, y)
2) By checking My = Nx to say M (x, y) + N (x, y) y 0 = 0 is exact.
3) Since M (x, y) + N (x, y) y 0 = 0 is exact, then there exists some φ(x, y) such that
φx (x, y) = M (x, y),
and φy (x, y) = N (x, y).
4) Since φx (x, y) = M (x, y), then
Z
φ(x, y) =
M (x, y) dx + f (y).
5) Since φy (x, y) = N (x, y), take the partial derivative with respect to y on the both sides of φ(x, y) =
Z
M (x, y) dx + f (y), then
Z
N (x, y) = φy (x, y) =
My (x, y) dx + f 0 (y).
6) Compute f (y), that is,
Z Z
f (y) =
N (x, y) − My (x, y) dx dy.
7) Write down the solution:
Z
φ(x, y) =
My (x, y) dx +
Z Z
N (x, y) − My (x, y) dx dy = C.
Example 1. Solve y(3x2 y + y 2 ) + y(2x3 + 3xy) y 0 = 0.
Let M (x, y) = y(3x2 y + y 2 ) = 3x2 y 2 + y 3 and N (x, y) = y(2x3 + 3xy) = 2x3 y + 3xy 2 , then
∂M
(x, y) = 6x2 y + 3y 2 ,
∂y
and
∂N
(x, y) = 6x2 y + 3y 2 .
∂x
So y(3x2 y + y 2 ) + y(2x3 + 3xy) y 0 = 0 is exact. That is, there exists some φ(x, y) such that
φx (x, y) = y(3x2 y + y 2 ),
and φy (x, y) = y(2x3 + 3xy).
1
2
MINGFENG ZHAO
Since φx (x, y) = ey (3x2 y + y 2 ), then
Z
φ(x, y) =
y(3x2 y + y 2 ) dx + f (y) = x3 y 2 + xy 3 + f (y).
Take the partial derivative with respect to y, since φy (x, y) = y(2x3 + 3xy), then
y(2x3 + 3xy) = φy (x, y) = 2x3 y + 3xy 2 + f 0 (y).
So f 0 (y) = 0, then we can take f (y) = 0. So the solution to y(3x2 y + y 2 ) + y(2x3 + 3xy) y 0 = 0 is:
x3 y 2 + xy 3 = C .
Question 1. How can we solve for the general differential equation M (x, y) + N (x, y) y 0 = 0 which may not be exact?
In general, consider differential equation:
M (x, y) + N (x, y) y 0 = 0.
Multiply some function µ(x, y) on both sides of equation:
[µM ] + [µN ] y 0 = 0.
If the above equation is exact (in this case, we call µ integrating factor), that is,
∂(µM )
∂(µN )
=
.
∂y
∂x
By the product rule, we get
µMy + M µy = µNx + N µx .
We have the following two special cases:
I. If µ(x) is a function of x, tha is, µy ≡ 0, then
µMy = µNx + N µx .
That is, µ satisfies
µx = µ ·
My − Nx
.
N
In this case, we only need
M y − Nx
is a function of x.
N
Hence
µ=e
R
My −Nx
N
dx
.
LECTURE 7: INTEGRATING FACTOR AND EULER’S METHOD
3
II. If µ(y) is a function of y, tha is, µx ≡ 0, then
µMy + M µy = µNx .
That is, µ satisfies
Nx − My
.
M
µy = µ ·
In this case, we only need
Nx − M y
is a function of y.
M
So we get
µ=e
R
Nx −My
M
dy
.
Remark 1. In homework and exams, we only consider µ is a function of only x. In this case, to solve the general
differential equation: M (x, y) + N (x, y) y 0 = 0:
My (x, y) − Nx (x, y)
.
N (x, y)
2) Understand the meaning of the integrating factor µ:
1) Compute My (x, y), Nx (x, y) and
[µM ] + [µN ] y 0 = 0
is exact, that is,
∂(µN )
∂(µM )
=
.
∂y
∂x
3) Take the integrating factor:
R
µ(x) = e
My −Nx
N
dx
4) Solve the exact equation: (µM ) + (µN ) y 0 = 0.
Example 2. Solve
Let M (x, y) =
y2
+ 2yex + (y + ex )y 0 = 0.
2
y2
+ 2yex and N (x, y) = y + ex , then
2
My (x, y) = y + 2ex ,
and Nx (x, y) = ex .
Then
My − Nx = y + 2ex − ex = y + ex = N.
So
M y − Nx
=1
N
is a function of x.
Then we can take the integrating factor
µ(x) = e
R
x dx
= ex .
4
MINGFENG ZHAO
That is, we get an exact equation:
e
x
y2
+ 2yex
2
y2
+ 2yex ,
2
+ ex (y + ex ) y 0 = 0.
Then there exists some φ(x, y) such that
φx = e x
x
Since φx = e
and φy = ex (y + ex ).
y2
x
+ 2ye , then
2
2
Z
y
y2
x
x
φ= e
+ 2ye
dx + f (y) = ex + ye2x + f (y).
2
2
Take the partial derivative with respect to y, since φy = N , then we get
ex (y + ex ) = N (x, y) = φy (x, y) = yex + e2x + f 0 (y).
Then
f 0 (y) = 0.
So we can take f (y) = 0. Hence the solution to
y2
+ 2yex + (y + ex )y 0 = 0 is:
2
y2 x
e + e2x y = C .
2
Example 3. Solve y 0 + p(x)y = q(x).
Rewrite the equation:
[p(x)y − q(x)] + y 0 = 0.
Let M (x, y) = p(x)y − q(x) and N (x, y) = 1, then
My = p(x),
and Nx = 0.
So
M y − Nx
= p(x)
N
is a function of x.
Then we can take the integrating factor
µ(x) = e
R
p(x) dx
.
That is, we get an exact equation:
e
R
p(x) dx
[p(x)y − q(x)] + e
R
p(x) dx
y 0 = 0.
LECTURE 7: INTEGRATING FACTOR AND EULER’S METHOD
5
Then there exists some φ(x, y) such that
φx = e
Since φy = e
R
p(x) dx
R
p(x) dx
[p(x)y − q(x)],
and φy = e
R
p(x) dx
.
, then
R
p(x) dx
φ(x, y) = ye
+ g(x).
Take the partial derivative with respect to x, we get
R
e
p(x) dx
R
[p(x)y − q(x)] = φx = ye
p(x) dx
· p(x) + g 0 (x).
So
g 0 (x) = −q(x)e
R
p(x) dx
.
That is,
Z
g(x) = −
R
p(x) dx
.
R
p(x) dx
= 0.
q(x)e
So we get
R
ye
p(x) dx
Z
−
q(x)e
Hence, the general solution to y 0 + p(x)y = q(x) is:
y=e
−
R
p(x) dx
Z
·
R
q(x)e
p(x) dx
dx .
Numerical methods: Euler’s method
Consider the initial value problem:
y 0 = f (x, y),
y(x0 ) = y0 .
Euler’s method: Given the initial condition y(x0 ) = y0 and the step size h, compute the point (xk+1 , yk+1 ) from the
previous point (xk , yk ) as follows:
1) Use the differential equation to compute the slope f (xk , yk ).
2) Calculate the next point (xk+1 , yk+1 ) using the formula
xk+1
= xk + h
yk+1
= yk + hf (xk , yk )
6
MINGFENG ZHAO
Let xk = x0 + kh, and
x1 = x0 + h
y1 = y0 + hf (x0 , y0 )
x2 = x1 + h
y2 = y1 + hf (x1 , y1 )
x3 = x2 + h
..
.
y3 = y2 + hf (x2 , y2 )
..
.
xk+1 = xk + h
..
.
yk+1 = yk + hf (xk , yk )
..
.
Example 4. Use Euler’s method to approximate y(3), where y 0 =
y2
, y(0) = 1.
3
If h = 1, then
k
xk
yk
f (xk , yk )
0
0
1
1
3
1
1
4
3
16
27
2
2
52
27
2704
2187
3
3
6916
2187
≈ 3.16232
dy
= 2y − 1, y(0) = 1, h = 0.1. Approximate the solution over the interval 0 ≤ t ≤ 0.5.
dx
We should compute 5 interactions of Euler’s method:
Example 5.
k
xk
yk
f (xk , yk )
0
0
1
1
1
0.1 1.1
1.2
2
0.2 1.22
1.44
3
0.3 1.364
1.73
4
0.4 1.537
2.07
5
0.5 1.744
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca