Chapter 3.6:
equations
Non-homogeneous
So far, we have looked at homogeneous equations
L[y] = y 00 + p(t)y 0 + q(t)y = 0.
Homogeneous means that the right side is zero.
Linear homogeneous equations satisfy the superposition principle: sums of solutions are solutions.
We now look at non-homogeneous equations:
L[y] = y 00 + p(t)y 0 + q(t)y = g(t),
where the RHS is not necessarily zero.
1
The New Issue
It is: Find a solution of an inhomogeneous
equation, say Y (t):
L[Y ] = Y 00 + p(t)Y 0 + q(t)Y = g(t).
The general solution is then:
y(t) = Y (t) + c1y1(t) + c2y2(t)
where y1, y2 are a fundamental set of solutions
of the homogeneous equation L[y] = 0.
Reason:
L[y] = L[Y ] + L[c1y1 + c2y2] = L[Y ] + 0 = g.
2
General solution
Let’s make sure that
y(t) = Y (t) + c1y1(t) + c2y2(t)
is the general solution of L[y] = g. Suppose
you have another, say yother . Then
L[y] = g = L[yother ]
=⇒
L[y − yother ] = 0
=⇒
y − yother = c1y1 + c2y2,
for some c1, c2, since y−yother solves the homogeneous equation, and the RHS is the general
solution.
3
How to solve L[Y ] = g
The first method is the method of undetermined coefficients = MUC. It only applies when
you know in advance what kind of solution the
equation will have. In the next chapter, a more
systematic method, variation of parameters,
will be used.
But for special equations such as constant coefficient equations, where the RHS g is an exponential function, MUC is quicker.
This motivates learning how to guess the type
of solution.
4
How to solve L[Y ] = g for CC equations
Example: Find a solution of y 00 −3y 0 −4y = 3e2t.
Key fact (which we already know): if Ly =
ay 00 + by 0 + cy, then
L[ert] = (ar2 + br + c)ert.
Thus, in our problem,
L[ert] = (r2 − 3r − 4)ert.
When solving the homogeneous equation, we
wanted the RHS = 0 so we chose r to be a
solution of the characteristic equation. But
now, we want the RHS to be 3e2t. Clearly, we
have to pick r = 2.
5
Choosing the multiple
We can’t always solve with Y = e2t but we can
try Y = Ae2t and solve for A. Indeed,
L[Ae2t] = A(22 − 3 · 2 − 4)e2t
= 3e2t ⇐⇒ −6A = 3 ⇐⇒ A = − 1
2.
2t . Since it is
Thus, one solution is Y = − 1
e
2
only one solution, we call it a particular solution.
To find the general solution, we need to add
the general solution of the homogeneous problem. (We won’t do it , because that was last
chapter).
6
Harder example
Find a solution of: y 00 − 3y 0 − 4y = 2 sin t.
d sin t = cos t, d cos t =
This is harder because dt
dt
− sin t. So sines and cosines are not quite preserved by the derivative.
So we have to try:
Y = A sin t + B cos t.
Functions of this kind are preserved by taking
a derivative.
7
Choosing the coefficients
Write: L[y] = y 00 − 3y 0 − 4y.
We claim that there exist constants A, B so
that
L[A sin t + B cos t] = 2 sin t.
A bit of computation shows:
L[A sin t + B cos t] = A(− sin t − 3 cos t − 4 sin t)
+ B(− cos t + 3 sin t − 4 cos t).
So we need:
A(− sin t − 3 cos t − 4 sin t)
+B(− cos t + 3 sin t − 4 cos t) = 2 sin t.
8
Choosing the coefficients
Equivalently,
(−A − 4A + 3B) sin t
+(−3A − B − 4B) cos t = 2 sin t
⇐⇒ −5A + 3B = 2,
−3A − 5B = 0.
Thus,
5 B = 2 ⇐⇒ B = 3 .
A = −5
B
=⇒
3B
+
5
3
3
17
5 and Y = − 5 sin t + 3 cos t.
So A = − 17
17
17
9
A disease
Unfortunately, this method does not work if
the RHS g is a solution of the homogeneous
equation L[g] = 0 on the LHS. For instance,
y 00 + y = sin t.
You cannot just try C1 cos t + C2 sin t since the
LHS will kill it.
The cure is, as in reduction of order, to multiply by t. Try
y = C1t cos t + C2t sin t.
Only do this when the RHS is a homogeneous
solution!
If you do this, you will find that terms with a
t in front are killed. So what remains is the
equation
−2C1 sin t + 2C2 cos t = sin t.
The factor of 2 comes from (yt)00 = 2y 0 + ty 00.
10
Example
Solve the initial value problem:
y 00 + 4y = sin 2t, y(0) = 0, y 0(0) = 1.
The RHS is a solution of the homogeneous
equation y 00+4y = 0 so we need to try At cos 2t+
Bt sin 2t. The equation becomes:
−1
−4A sin 2t+4B cos 2t = sin 2t =⇒ A =
, B = 0.
4
The general solution is:
1
y = C1 cos 2t + C2 sin 2t − t cos 2t.
4
Then, y(0) = C1 = 0; y 0(0) = −2C2 − 1
4 = 1.
So the solution is
5
1
y = − sin 2t − t cos 2t.
8
4
11
Repeated root canal
Yet a worse disease occurs if the LHS has repeated roots and the RHS is a solution of the
homogeneous equation. For instance:
L[y] = y 00 + 2y 0 + 1 = e−t.
There is no point trying y = te−t because it
is also a homogeneous solution. So the next
step is to try y = t2e−t. L kills terms with t2
or t. The only remaining term is 2e−t. So a
solution is:
1
y = t2e−t.
2
12
What else can the RHS be?
So far we have solved L[y] = g when Ly =
ay 00 + by 0 + cy and where g is an exponential or
a cosine or a sine.
We can also solve by undetermined coefficients
if the RHS is a polynomial.
This is because derivatives of polynomials are
polynomials. The degree goes down with each
derivative. If the RHS is a polynomial of degree
n, and if c 6= 0, you can use a polynomial of
degree n. E.g.:
L[y] = y 00 + y = t2 + 2.
Try y = at2 + bt + c. Then
L[at2 + bt + c] =
⇐⇒
2a + at2 + bt + c = t2 + 2
2a + c = 2, b = 0, a = 1.
13
Another example
Example: y 00 − 3y 0 = 4t2 − 1.
We can’t use a polynomial of degree 2 since
the LHS will then be of degree one. Try y =
polynomial of degree 3. Thus, y = At3 + Bt2 +
Ct. We now have 3 undetermined coefficients
(A, B, C).
Plugging in, we get:
(6At + 2B) − 3(3At2 + 2tB + C) = 4t2 − 1.
Match coefficients of like powers of t:
−9A = 4, 6A − 6B = 0, −3C = −1.
14
What else can the RHS be?
A nice thing is that we can easily solve when
g = g1 + g2 if we can solve separately for g1
and for g2. Indeed, if
L[Y1] = g1, L[Y2] = g2
=⇒ L[Y1 + Y2] = g1 + g2 = g.
This is the superposition principle again (i.e.
linearity of L).
So when L = aD2 + bD + c, our inventory of
g’s now includes: sums of exponentials, sines,
cosines and polynomials.
15
What else can the RHS be?
If we can solve when the RHS is an exponential,
then we can surely solve when the exponential
is a complex one. That means we can solve if
the RHS g = eλt cos t or g = eλt sin t or a sum
of these.
One can also solve when the RHS is a polynomial times an exponential.
16
Examples
L[y] = y 00 − 4y = 2t + e2t.
We solve one term at a time.
First, we find y1 such that L[y1] = 2t. We can
use at + b. Then L[y] = −4at − 4b and we get
a = −1
2 , b = 0.
Then we find y2 such that L[y2] = e2t. This is
a solution of the homogeneous problem so we
must use Ate2t. We get
L[Ate2t] = 4Ae2t = e2t =⇒ A =
1
.
4
1 te2t.
t
+
The solution is: y = − 1
2
4
17