Download Integrating Factors and Reduction of Order

Integrating
Factors and
Reduction of
Order
Math 240
Integrating
factors
Reduction of
order
Integrating Factors and Reduction of Order
Math 240 — Calculus III
Summer 2015, Session II
Monday, August 3, 2015
Integrating
Factors and
Reduction of
Order
Agenda
Math 240
Integrating
factors
Reduction of
order
1. Integrating factors
2. Reduction of order
Integrating
Factors and
Reduction of
Order
Introduction
Math 240
Integrating
factors
Reduction of
order
The reduction of order technique, which applies to
second-order linear differential equations, allows us to go
beyond equations with constant coefficients, provided that we
already know one solution.
If our differential equation is
y 00 + a1 (x)y 0 + a2 (x)y = F (x),
and we know the solution, y1 (x), to the associated
homogeneous equation, this method will furnish us with
another, independent solution.
To accomplish the process, we will make use of integrating
factors.
Integrating
Factors and
Reduction of
Order
Integrating factors
Math 240
Integrating
factors
Reduction of
order
Integrating factors are a technique for solving first-order linear
differential equations, that is, equations of the form
dy
a(x)
+ b(x)y = r(x).
dx
Assuming a(x) 6= 0, we can divide by a(x) to put the equation
in standard form:
dy
+ p(x)y = q(x).
dx
The main idea is that the left-hand side looks almost like the
result of the product rule for derivatives. If I(x) is another
function then
d
dy
dI
(Iy) = I
+
y.
dx
dx dx
The standard form equation is missing an I in front of
let’s multiply it by I.
dy
dx ,
so
Integrating
Factors and
Reduction of
Order
Integrating factors
Math 240
Integrating
factors
Reduction of
order
When we multiply our equation by I, we get
dy
I
+ Ip(x)y = Iq(x),
dx
d
(Iy), we need to have
so in order for the left-hand side to be dx
dI
= p(x)I.
dx
Rearranging this into
dI
= p(x) dx,
I
we can solve:
R
I(x) = c1 e p(x) dx .
Since we only need one function I, let’s set c1 = 1.
Integrating
Factors and
Reduction of
Order
Integrating factors
Math 240
Integrating
factors
Reduction of
order
Using this I, we rewrite our equation as
d
(Iy) = q(x)I,
dx
then integrate and divide by I to get
Z
1
y(x) =
q(x)I dx + c .
I
Our I is called an integrating factor because it is something
we can multiply by (a factor) that allows us to integrate.
Integrating
Factors and
Reduction of
Order
Math 240
Example
Find a solution to
y 0 + xy = xex
2 /2
.
Integrating
factors
Reduction of
order
1. Find the integrating factor
I(x) = e
R
x dx
= ex
2 /2
.
2. Multiply it into the original equation:
d x2 /2 2
2
2
e
y = ex /2 y 0 + xex /2 y = xex .
dx
3. Integrate both sides:
1 2
y = ex + c.
2
4. Divide by I to find the solution
−x2 /2 1 x2
y(x) = e
e +c .
2
ex
2 /2
Integrating
Factors and
Reduction of
Order
Math 240
Integrating
factors
Reduction of
order
Example
Solve, for x > 0, the equation
xy 0 + 2y = cos x.
1. Write the equation in standard form:
2
cos x
y0 + y =
.
x
x
2. An integrating factor is
I(x) = e2 ln x = x2 .
3. Multiply by I to get
d 2
(x y) = x cos x.
dx
4. Integrate and divide by x2 to get
x sin x + cos x + c
y(x) =
.
x2
Integrating
Factors and
Reduction of
Order
Reduction of order
Math 240
Integrating
factors
Reduction of
order
We now turn to second-order equations
y 00 + a1 (x)y 0 + a2 (x)y = F (x).
We know that the general solution to such an equation will
look like
y(x) = c1 y1 (x) + c2 y2 (x) + yp (x).
Suppose that we know y1 (x). We will guess the solution
y(x) = u(x)y1 (x). Plugging it into our original equation yields
u00 y1 + u0 2y10 + a1 (x)y1 = F (x).
If we let w = u0 then we have reduced our second-order
equation to the first-order
0
2y1
F (x)
0
w +
+ a1 w =
.
y1
y1
Integrating
Factors and
Reduction of
Order
Reduction of order
Math 240
Integrating
factors
Reduction of
order
We may solve
2y10
F (x)
w +
+ a1 w =
y1
y1
using the integrating factor technique:
0
I(x) = y12 (x)e
and
1
w(x) =
I(x)
Z
x
Rx
a1 (s) ds
I(s)F (s)
c1
ds +
.
y1 (s)
I(x)
Then integrate w to find u:
Z x
Z t
Z
1
I(s)F (s)
u(x) =
ds dt + c1
I(t)
y1 (s)
x
1
ds + c2 .
I(s)
Integrating
Factors and
Reduction of
Order
Reduction of order
Math 240
Integrating
factors
Reduction of
order
Finally, we get
Z
x
1
ds + c2 y1 (x)
I(s)
Z x
Z t
1
I(s)F (s)
+ y1 (x)
ds dt.
I(t)
y1 (s)
Using F = 0 gives us the two fundamental solutions
Z x
1
y(x) = y1 (x) and y(x) = y1 (x)
ds.
I(s)
And using c1 = c2 = 0, we get a particular solution
Z x
Z t
1
I(s)F (s)
yp (x) = y1 (x)
ds dt.
I(t)
y1 (s)
y(x) = u(x)y1 (x) = c1 y1 (x)
Integrating
Factors and
Reduction of
Order
Math 240
Integrating
factors
Reduction of
order
Example
Determine the general solution to
xy 00 − 2y 0 + (2 − x)y = 0,
given that one solution is y1 (x) =
x > 0,
ex .
1. Set up the equation for w:
2(x − 1)
w = 0.
w0 +
x
2. Solve for w:
w(x) = c1 x2 e−2x .
3. Integrate to find
Z
1
u(x) = w(x) dx + c2 = − c1 e−2x (1 + 2x + 2x2 ) + c2 .
4
4. Multiply by y1 for the general solution:
y(x) = c1 e−x (1 + 2x + 2x2 ) + c2 ex .
Integrating
Factors and
Reduction of
Order
Math 240
Integrating
factors
Reduction of
order
Example
Determine the general solution to
x2 y 00 + 3xy 0 + y = 4 ln x,
x > 0,
by first finding solutions to the associated homogeneous
equation of the form y(x) = xr .
1. Find y1 (x) = x−1 .
2. Put the equation in standard form by dividing by x2 :
y 00 + 3x−1 y 0 + x−2 y = 4x−2 ln x.
3. Set up the equation w0 + x−1 w = 4x−1 ln x.
4. Find w(x) = 4(ln x − 1) + c1 x−1 .
5. Then u(x) = 4x(ln x − 2) + c1 ln x + c2 .
6. Multiply by y1 (x) = x−1 :
y(x) = 4(ln x − 2) + c1 x−1 ln x + c2 x−1 .