# Toronto Math Forum

## MAT244-2018S => MAT244--Tests => Term Test 1 => Topic started by: Victor Ivrii on February 13, 2018, 09:25:20 PM

Title: P-3
Post by: Victor Ivrii on February 13, 2018, 09:25:20 PM
(a) Find the general solution for equation
\begin{equation*}
y''(t)+y'(t)-2y(t)=-6+9 e^{-2t} .
\end{equation*}

(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Title: Re: P-3
Post by: David Chan on February 13, 2018, 10:50:31 PM
(a) The characteristic equation for $\mathcal{L}[y] = y'' + y' - 2y$ is $r^2 + r - 2 = 0$ which has roots $r_1 = 1$ and $r_2 = -2$.  Thus, the general solution to the corresponding homogeneous equation $\mathcal{L}[y_c] = 0$ is $$y_c = c_1e^t + c_2e^{-2t}$$
We now look for a particular solution to the non-homogeneous equation $\mathcal{L}[Y] = -6 + 9e^{-2t}$ by method of undetermined coefficients.
First, we look for $Y_1 = A$ such that $\mathcal{L}[Y_1] = -6$.  Differentiating, we get $Y_1' = Y_1'' = 0$.  Plugging these back into the differential equation, we get $$-2A = -6 \implies A = 3$$
Note that $e^{-2t}$ is already part of the homogeneous solution, so instead, we look for $Y_2 = Bte^{-2t}$ such that $\mathcal{L}[Y_2] = 9e^{-2t}$.  Differentiating, we get $Y_2' = Be^{-2t} -2Bte^{-2t}$ and $Y_2'' = 4Bte^{-2t} - 4Be^{-2t}$.  Plugging these back into the differential equation, we get $$4Bte^{-2t} - 4Be^{-2t} + Be^{-2t} -2Bte^{-2t} - 2Bte^{-2t} = -3Be^{-2t} = 9e^{-2t} \quad \implies\quad B = -3$$
So, we have $Y_1 = 3$ and $Y_2 = -3te^{-2t}$.  Thus, the general solution to the differential equation is $$\boxed{y = c_1e^t + c_2e^{-2t} - 3te^{-2t} + 3}$$

(b) Note that $y'(t) = c_1 e^t - 2 c_2 e^{-2 t} + 6 t e^{-2 t} - 3 e^{-2 t}$.  Given the initial conditions $y(0) = 0$ and $y'(0) = 0$, we find that $$0 = c_1 + c_2 + 3$$ $$0 = c_1 - 2c_2 - 3$$
Solving, this system of equations, we get $c_1 = -1$ and $c_2 = -2$, so our solution is $$\boxed{y = -e^t - 2e^{-2t} - 3te^{-2t} + 3}$$