So far, we have looked at homogeneous equations

Chapter 3.6: Non-homogeneous equations

So far, we have looked at homogeneous equa- tions

L[y] = y⁰⁰ + p(t)y⁰ + q(t)y = 0.

Homogeneous means that the right side is zero.

Linear homogeneous equations satisfy the su- perposition principle: sums of solutions are so- lutions.

We now look at non-homogeneous equations:

L[y] = y⁰⁰ + p(t)y⁰ + q(t)y = g(t), where the RHS is not necessarily zero.

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.

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 y_other. Then

L[y] = g = L[y_other] =⇒ L[y − y_other] = 0

=⇒ y − y_other = c₁y₁ + c₂y₂, for some c₁, c₂, since y−y_other solves the homo-

geneous equation, and the RHS is the general solution.

How to solve L[Y ] = g

The first method is the method of undeter- mined 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 co- efficient equations, where the RHS g is an ex- ponential function, MUC is quicker.

This motivates learning how to guess the type of solution.

How to solve L[Y ] = g for CC equa- tions

Example: Find a solution of y⁰⁰−3y⁰−4y = 3e^2t. Key fact (which we already know): if Ly = ay⁰⁰ + by⁰ + cy, then

L[e^rt] = (ar² + br + c)e^rt. Thus, in our problem,

L[e^rt] = (r² − 3r − 4)e^rt.

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 3e^2t. Clearly, we have to pick r = 2.

Choosing the multiple

We can’t always solve with Y = e^2t but we can try Y = Ae^2t and solve for A. Indeed,

L[Ae^2t] = A(2² − 3 · 2 − 4)e^2t

= 3e^2t ⇐⇒ −6A = 3 ⇐⇒ A = −¹₂.

Thus, one solution is Y = −¹₂e^2t. Since it is only one solution, we call it a particular solu- tion.

To find the general solution, we need to add the general solution of the homogeneous prob- lem. (We won’t do it , because that was last chapter).

Harder example

Find a solution of: y⁰⁰ − 3y⁰ − 4y = 2 sin t.

This is harder because _dt^d sin t = cos t, _dt^d cos t =

− sin t. So sines and cosines are not quite pre- served 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.

Choosing the coefficients

Write: L[y] = y⁰⁰ − 3y⁰ − 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.

Choosing the coefficients

Equivalently,

(−A − 4A + 3B) sin t

+(−3A − B − 4B) cos t = 2 sin t

⇐⇒ −5A + 3B = 2, −3A − 5B = 0.

Thus,

A = −⁵₃B =⇒ 3B + 5⁵₃B = 2 ⇐⇒ B = ₁₇³ . So A = −₁₇⁵ and Y = −₁₇⁵ sin t + ₁₇³ cos t.

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,

y⁰⁰ + 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 mul- tiply 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.

Example

Solve the initial value problem:

y⁰⁰ + 4y = sin 2t, y(0) = 0, y⁰(0) = 1.

The RHS is a solution of the homogeneous equation y⁰⁰+4y = 0 so we need to try At cos 2t+

Bt sin 2t. The equation becomes:

−4A sin 2t+4B cos 2t = sin 2t =⇒ A = −1

4 , B = 0.

The general solution is:

y = C₁ cos 2t + C₂ sin 2t − 1

4t cos 2t.

Then, y(0) = C₁ = 0; y⁰(0) = −2C₂ − ¹₄ = 1.

So the solution is y = −5

8 sin 2t − 1

4t cos 2t.

Repeated root canal

Yet a worse disease occurs if the LHS has re- peated roots and the RHS is a solution of the homogeneous equation. For instance:

L[y] = y⁰⁰ + 2y⁰ + 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:

y = 1

2t²e^−t.

What else can the RHS be?

So far we have solved L[y] = g when Ly = ay⁰⁰ + by⁰ + 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] = y⁰⁰ + 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.

Another example

Example: y⁰⁰ − 3y⁰ = 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.

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.

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 = e^λt cos t or g = e^λt sin t or a sum of these.

One can also solve when the RHS is a polyno- mial times an exponential.

Examples

L[y] = y⁰⁰ − 4y = 2t + e^2t. 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 = ⁻¹₂ , b = 0.

Then we find y₂ such that L[y₂] = e^2t. This is a solution of the homogeneous problem so we must use Ate^2t. We get

L[Ate^2t] = 4Ae^2t = e^2t =⇒ A = 1 4. The solution is: y = −¹₂t + ¹₄te^2t.