Toronto Math Forum
MAT2442018S => MAT244Tests => Term Test 1 => Topic started 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 $y(0)=0$, $y'(0)=0$.

(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 nonhomogeneous 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 $$\boxed{y = e^t  2e^{2t}  3te^{2t} + 3}$$