### Author Topic: P-4  (Read 2525 times)

#### Victor Ivrii

• Elder Member
• Posts: 2599
• Karma: 0
##### P-4
« on: February 13, 2018, 09:25:55 PM »
Find the general solution for equation
\begin{equation*}
y''(t)+2y'(t)+2y(t)=-e^{-t}+ 10\cos(t).
\end{equation*}

#### David Chan

• Jr. Member
• Posts: 13
• Karma: 8
##### Re: P-4
« Reply #1 on: February 13, 2018, 11:15:40 PM »
The characteristic equation for $\mathcal{L}[y] = y'' + 2y' + 2y$ is $r^2 + 2r + 2 = 0$ which has roots $r = -1 \pm i$.  Thus, the general solution to the corresponding homogeneous equation $\mathcal{L}[y_c] = 0$ is $$y_c = c_1e^{-t}\cos(t) + c_2e^{-t}\sin(t)$$
We now look for a particular solution to the non-homogeneous equation $\mathcal{L}[y] = -e^{-t} + 10\cos(t)$ by method of undetermined coefficients.
First, we look for $Y_1 = Ae^{-t}$ such that $\mathcal{L}[Y_1] = -e^{-t}$.  Differentiating, we get $Y_1' = -Ae^{-t}$ and $Y_1'' = Ae^{-t}$.  Plugging these back into the differential equation, we get $$Ae^{-t} - 2Ae^{-t} + 2Ae^{-t} = Ae^{-t} = -e^{-t} \quad \implies \quad A = -1$$
Next, we look for $Y_2 = B\cos(t) + C\sin(t)$ such that $\mathcal{L}[Y_2] = 10\cos(t)$.  Differentiating, we get $Y_2' = -B\sin(t) + C\cos(t)$ and $Y_2'' = -B\cos(t) - C\sin(t)$.  Plugging these back into the differential equation, we get $$-B\cos(t) - C\sin(t) - 2B\sin(t) + 2C\cos(t) + 2B\cos(t) + 2C\sin(t) = (B + 2C)\cos(t) + (-2B + C)\sin(t)$$ $$\implies \begin{cases} \ \ B + 2C &= 10 \\ -2B + C &= 0 \end{cases} \quad \implies \quad B = 2,\ C = 4$$
So, we have $Y_1 = -e^{-t}$ and $Y_2 = 2\cos(t) + 4\sin(t)$.  Thus, the general solution to the differential equation is $$\boxed{y = c_1e^{-t}\cos(t) + c_2e^{-t}\sin(t) -e^{-t} + 2\cos(t) + 4\sin(t)}$$
« Last Edit: February 14, 2018, 12:09:47 AM by David Chan »