# Solving Nonhomogeneous Linear ODEs using Annihilators

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My previous Insights article, Solving Homogeneous Linear ODEs using Annihilators, discussed several examples of homogeneous differential equations, equations of the form F(y, y’, y”, …) = 0. In this Insights article we will look at equations of the form F(y, y’, y”, …) = g(t), for certain functions g.

Ex. 1: y” – 4y’ + 3y = 5

For this example, we note two facts:

1. The left side can be rewritten as $(D^2 – 4D + 3)y$.
2. The derivative of 5 = 0; that is, D(5) = 0.

If I apply the D operator to both sides, I get:

$$D(D^2 – 4D – 3)y = D(5) = 0$$

or, in factored form:

$$D(D – 3)(D – 1)y = 0$$

By “annihilating” the right side, I have converted the nonhomogeneous differential equation y” – 4y’ + 3y = 5 into a homogeneous differential equation, albeit of a higher order.

The characteristic equation of the new equation can be read off from the operator notation: $r(r – 3)(r – 1) = 0$. The roots are r = 0, r = 3, and r = 1, so a fundamental solution set is $\{e^{0t} = 1, e^{3t}, e^t\}$.

The general solution for the third-order homogeneous equation is $y(t) = A \cdot 1 + c_1e^{3t} + c_2e^t$. You might wonder why I have different coefficients for the three fundamental solutions. The reason is that I want to keep the one (A) that came from the nonhomogeneous problem separate from the other two ($c_1$ and $c_2$).

Differentiating the solution twice, I get:

$y(t) = A \cdot 1 + c_1e^{3t} + c_2e^t$

$y'(t) = 3c_1e^{3t} + c_2e^t$

$y”(t) = 9c_1e^{3t} + c_2e^t$

Substituting back into the original differential equation, I get

$9c_1e^{3t} + c_2e^t – 4(3c_1e^{3t} + c_2e^t) + 3(A + c_1e^{3t} + c_2e^t) = 5$

$\Rightarrow (9c_1 – 12c_1 + 3c_1)e^{3t} + (c_2 – 4c_2 + 3 c_2)e^t + 3A = 5$

$\Rightarrow A = \frac 5 3$

The general solution to the original differential equation is $y(t) = \frac 5 3 + c_1e^{3t} + c_2e^t$.

Note: If initial conditions had been given in this problem, we could find the constants $c_1$ and $c_2$.

Ex. 2: $y” – 4y’ + 3y = e^{2t}$

This equation can be rewritten as $(D^2 – 4D + 3)y = e^{2t}$. Recall that the (D – 2) operator annihilates $e^{2t}$. That is, $(D – 2)e^{2t} = 0$.

If we apply the (D – 2) operator to both sides, we get $(D – 2)(D^2 – 4D + 3)y = (D – 2)e^{2t} = 0$, which is a homogeneous equation, of third order.

Factoring the operators, we have $(D – 2)(D – 3)(D – 1)y = 0$, so a fundamental solution set is $\{e^{2t}, e^{3t}, e^t\}$. Note that the characteristic equation of the third-order, homogeneous equation is (r – 2)(r – 3)(r – 1) = 0.

The general solution of the homogeneous equation is $y(t) = Ae^{2t} + c_1e^{3t} + c_2e^t$. If we substitute this into the original, nonhomogeneous equation, we get

$$(4A – 8A + 3A)e^{2t} + (9c_1 – 12c_1 + 3c_1)e^{3t} + (c_2 – 4c_2 + 3c_2)e^t = e^{2t}$$

$$\Rightarrow -Ae^{2t} = e^{2t} \Rightarrow A = -1$$

The general solution to the original nonhomogeneous equation is $y(t) = -e^{2t} + c_1e^{3t} + c_2e^t$.

Ex. 3:  $y” – 4y’ + 3y = e^{3t}$

Rewriting the equation in operator notation, we have $(D^2 – 4D + 3D)y = e^{3t}$, or $(D – 3)(D – 1)y = e^{3t}$

We notice that the right side can be annihilated with the (D – 3) operator.

If we apply this operator to both sides, we get $(D – 3)^2(D – 1) y = (D -3)e^{3t} = 0$, which is a third order, homogeneous equation.

One thing to notice is that, unlike the previous example, we now have a repeated factor $(D – 3)^2$. This means that our fundamental set of solutions will be $\{e^{3t}, te^{3t}, e^t\}$.

The general solution to the third order problem is $y(t) = c_1e^{3t} + Ate^{3t} + c_2e^t$. The A coefficient on the middle term is to mark it as stemming from the second-order nonhomogeneous problem.

If we substitute this function and its first two derivatives into the second-order equation, we get A = 1/2. (Verification left to the reader.)

The general solution is $y(t) = c_1e^{3t} + \frac 1 2 te^{3t} + c_2e^t$.

As before, if the problem included initial conditions, we could find the constants $c_1$ and $c_2$.

Ex. 4: $y” + 4y = te^{-t}$

In operator notation, the equation above is $(D^2 + 4)y = te^{-t}$. The annihilator of the function on the right side is $(D + 1)^2$. The reason for the exponent of 2 is because of the presence of t on the right side. The $D + 1$ operator annihilates $e^{-t}$, and the $(D + 1)^2$ operator annihilates $te^{-t}$.

Applying the $(D + 1)^2$ operator to both sides, we get $(D + 1)^2(D^2 + 4)y = (D + 1)^2(te^{-t} = 0$. What was a second-order nonhomogeneous equation is now a fourth order homogeneous equation.

With a fourth-order equation, we expect the fundamental solution set to consist of four linearly independent solutions: $\{e^{-t}, te^{-t}, \cos(2t), \sin(2t) \}$.

The general solution of the fourth-order equation is $y(t) = Ae^{-t} + Bte^{-t} + c_1\cos(2t) + c_2\sin(2t)$.

I haven’t mentioned this before, but the first two terms in the equation above represent a particular solution to the original nonhomogeneous problem. The last two terms represent the complementary solutions — they are solutions of the related homogeneous equation y” + 4y = 0.

Instead of taking the first two derivatives of y(t) above, we can save some work by taking the derivatives of $y_p(t) = Ae^{-t} + Bte^{-t}$, the particular solution to the nonhomogeneous equation, $y” + y = te^{-t}$.

$y_p(t)$ and its first two derivatives are:

$y_p(t) = Ae^{-t} + Bte^{-t}$

$y_p'(t) = -Ae^{-t} + Be^{-t} – Bte^{-t}$

$y_p”(t) = Ae^{-t} – Be^{-t} – Be^{-t} + Bte^{-t}$

Substituting into $y” + y = te^{-t}$, we get:

$y_p”(t) + y_p = (2A – 2B)e^{-t} + 2Bte^{-t}$, or

A = B = 1/2

The general solution of the equation of this example is $y(t) = \frac 1 2 e^{-t} + \frac 1 2 te^{-t} + c_1\cos(2t) + c_2\sin(2t)$.

Ex. 5: $y” + y = 2\cos(t)$

In operator notation, this equation is $(D^2 + 1)y = 2\cos(t)$. The annihilator of the right side is $(D^2 + 1)$. (The coefficient of 2 on the right side has no effect on the annihilator we choose. If an operator annihilates f(t), the same operator annihilates k*f(t), for any constant k.)

Applying the $(D^2 + 1)$ operator to both sides, we get $(D^2 + 1)(D^2 + 1)y = (D^2 + 1)(2\cos(t)) = 0$.

In slightly different form, we see that this is $(D^2 + 1)^2y = 0$, a fourth-order homogeneous equation whose characteristic equation has repeated roots.

For the set of fundamental solutions, we choose $\{ \cos(t), \sin(t), t\cos(t), t\sin(t)\}$. The general solution of the fourth-order homogeneous equation is $y(t) = c_1 \cos(t) + c_2 \sin(t) + At\cos(t) + Bt\sin(t)$. As before the terms with the coefficients of A and B make up the particular solution. The other two terms make up the complementary function, the general solution of the related homogeneous problem, y” + y = 0.

$y_p(t)$ and its first two derivatives are:

$y_p(t) = At\cos(t) + Bt\sin(t)$

$y_p'(t) = A\cos(t) – At\sin(t) + B\sin(t) + Bt\cos(t)$

$y_p”(t) = -A\sin(t) – A\sin(t) – At\cos(t) + B\cos(t) + B\cos(t) – Bt\sin(t) = -2A\sin(t) – At\cos(t) + 2B\cos(t) – Bt\sin(t)$

Substituting into the nonhomogeneous equation $y” + y = 2\cos(t)$, we can simplify things to $-2A\sin(t) + 2B\cos(t) = 2\cos(t) \Rightarrow A = 0$ and $B = 1$, so $y_p(t) = t\sin(t)$.

The general solution for the equation of this example is $y(t) = c_1\cos(t) + c_2\sin(t) + t\sin(t)$.

The following table lists several operators, together with the functions that they annihilate.

Operators and the functions they annihilate
OperatorFunction that is annihilated
$D, D^2, D^3, \dots$1, t, $t^2$, $t^3$, …
$D – k, (D – k)^2, (D – k)^3, \dots$$e^{kt}, te^{kt}, t^2e^{kt},$ …
$(D^2 + 1)$cos(t), sin(t)
$(D^2 + 1)^2$tcos(t), tsin(t)
$(D^2 + 4)$cos(2t), sin(2t)

Former college mathematics professor for 19 years; taught a variety of programming languages. Former technical writer for 15 years at a large software firm headquartered in Redmond, WA.
I enjoy traipsing around off-trail in Olympic National Park, as well as riding and tinkering with my four motorcycles.

5 replies
1. PWiz says:

Great entry! Please make some more math tutorials Mark!!

2. Mark44 says:

Great entry! Please make some more math tutorials Mark!!

Thank you! I have another one in mind coming soon, and probably some more after that.

3. PWiz says:

I think I found 2 small typos in the post. In ex 1, you said

Substituting back into the original differential equation, I get

$9c_1e^{3t}+c_2e^t–4(3c_1e^{3t}+c_2e^t)+3(A+c_1e^{3t}+c_2e^t)=5 ⇒(9c_1–12c_1+3c_1)e^{3t}+(c_2–4c_2+c_2)e^t+3A=5 ⇒A=5/3$

I think in the 2nd last step $(c_2–4c_2+3c_2)e^t$ should be over there instead of $(c_2–4c_2+c_2)e^t$.

In ex 4, you said

With a fourth-order equation, we expect the fundamental solution set to consist of four linearly independent solutions: {$e^{−t},te−t,cos(2t),sin(2t)$ }.

The 2nd solution should be $te^{-t}$ right?

4. Mark44 says:

I think I found 2 small typos in the post. In ex 1, you said

I think in the 2nd last step $(c_2–4c_2+3c_2)e^t$ should be over there instead of $(c_2–4c_2+c_2)e^t$.

In ex 4, you said

The 2nd solution should be $te^{-t}$ right?

Thanks for spotting these — I have fixed both of them.
Even though I looked through this stuff before publishing it, it’s still hard to catch everythin, especially when you’re working with LaTeX .