The Method of Annihilators Examples 1

# The Method of Annihilators Examples 1

On The Method of Annihilators page, we looked at an alternative way to solve higher order nonhomogeneous differential equations with constant coefficients apart from the method of undetermined coefficients. Consider a differential equation of the form:

(1)
\begin{align} \quad L(D)(y) = g(t) \end{align}

The procedure for solving this differential equation was straightforward. Like always, we first solved the corresponding homogeneous differential equation. We then rewrote the differential equation in terms of differential operators, and determined a differential operator $M(D)$ which annihilated $g(t)$, that is, $M(D)(g(t)) = 0$. We then apply this annihilator to both sides of the differential equation to get:

(2)

The result is a new differential equation that is now homogeneous. In solving this differential equation - we obtain a general solution for which we throw away terms that are linear combinations of the solution to the original corresponding homogeneous differential equation. We then have obtained a form for the particular solution $Y(t)$. We then differentiate $Y(t)$ as many times as necessary and plug it into the original differential equation and solve for the coefficients.

Lastly, as usual, we obtain the general solution to our higher order differential equation as:

(3)
\begin{align} \quad y(t) = y_h(t) + Y(t) \end{align}

We will now look at an example of applying the method of annihilators to a higher order differential equation.

## Example 1

Solve the differential equation $\frac{\partial^4 y}{\partial t^4} - 2 \frac{\partial^2 y}{\partial t} + y = e^t + \sin t$ using the method of annihilators.

The corresponding homogeneous differential equation is $\frac{\partial^4 y}{\partial t^4} - 2 \frac{\partial^2 y}{\partial t} + y$ and the characteristic equation is $r^4 - 2r^2 + 1 = (r^2 - 1)^2 = (r + 1)^2(r - 1)^2 = 0$. Therefore the characteristic equation has two distinct roots $r_1 = 1$ and $r_2 = -1$ - each with multiplicity $2$, and so the general solution to the corresponding homogeneous differential equation is:

(4)
\begin{align} \quad y_h(t) = Ae^{t} + Bte^{t} + Ce^{-t} + Dte^{-t} \end{align}

We now rewrite our differential equation in terms of differential operators as:

(5)
\begin{align} \quad (D + 1)^2(D - 1)^2(y) = e^t + \sin t \end{align}

The differential operator $(D - 1)$ annihilates $e^t$ since $(D - 1)(e^t) = D(e^t) - e^t = e^t - e^t = 0$. The differential operator $(D^2 + 1)$ annihilates $\sin t$ since $(D^2 + 1)(\sin t) = D^2(\sin t) + \sin t = -\sin t + \sin t = 0$. Applying the operator $(D^2 + 1)(D - 1)$ to both sides of the differential equation above gives us:

(6)
\begin{align} \quad (D + 1)^2(D - 1)^2(y) = e^t + \sin t \\ \quad (D^2 + 1)(D + 1)^2(D - 1)^3 (y) = (D^2 + 1)(D - 1)(e^t + \sin t) \\ \quad (D^2 + 1)(D + 1)^2(D - 1)^3 (y) = 0 \end{align}

The roots to the characteristic polynomial of the differential equation above are $r_1 = i$, $r_2 = -i$, $r_3 = -1$ (with multiplicity $2$), $r_4 = 1$ (with multiplicity $3$), and so the general solution to the differential equation above is:

(7)
\begin{align} \quad Y(t) = P \sin t + Q \cos t + Re^{-t} + Ste^{-t} + Ue^{t} + Vte^{t} + Wt^2e^{t} \end{align}

The terms $Re^{-t}$, $Ste^{-t}$, $Ue^{t}$, and $Vte^{t}$ are all contained in the linear combination of the corresponding homogeneous differential from the beginning of this example. Therefore, we discard them to get:

(8)
\begin{align} \quad Y(t) = P \sin t + Q \cos t + Wt^2 e^t \end{align}

We now need to differentiate $Y(t)$ four times to get:

(9)
\begin{align} \quad Y'(t) = P \cos t - Q \sin t + W(2t + t^2)e^t \end{align}
(10)
\begin{align} \quad Y''(t) = -P \sin t - Q \cos t + W(2 + 4t + t^2)e^t \end{align}
(11)
\begin{align} \quad Y'''(t) = -P \cos t + Q \sin t + W(6 + 6t + t^2)e^t \end{align}
(12)
\begin{align} \quad Y^{(4)} = P \sin t + Q \cos t + W(12 + 8t + t^2)e^t \end{align}

Plugging this into the original differential equation gives us:

(13)
\begin{align} \quad \quad \frac{\partial^4 y}{\partial t^4} - 2 \frac{\partial^2 y}{\partial t} + y = e^t + \sin t \\ \quad \quad \left [ P \sin t + Q \cos t + W(12 + 8t + t^2)e^t \right ] - 2 \left [ -P \sin t - Q \cos t + W(2 + 4t + t^2)e^t \right ] + \left [ P \sin t + Q \cos t + Wt^2 e^t \right ] = e^t + \sin t \\ \quad \quad \left ( P + 2P + P \right ) \sin t + \left ( Q + 2Q + Q \right ) \cos t + \left (12W - 4W \right ) e^t + \left (8W - 8W \right )te^t + \left ( W - 2W + W \right ) t^2 e^t = e^t + \sin t \\ \quad 4P \sin t + 4Q \cos t + 8W e^t = e^t + \sin t \end{align}

From the equation above, we see that $P = \frac{1}{4}$, $Q = 0$, and $W = \frac{1}{8}$. Therefore a particular solution to our differential equation is:

(14)
\begin{align} \quad Y(t) = \frac{1}{4} \sin t + \frac{1}{8}t^2 e^t \end{align}

The general solution to our original differential equation is therefore:

(15)
\begin{align} \quad y(t) = y_h(t) + Y(t) \\ \quad y(t) = Ae^{t} + Bte^{t} + Ce^{-t} + Dte^{-t} + \frac{1}{4} \sin t + \frac{1}{8}t^2 e^t \end{align}