Toronto Math Forum
MAT2442018S => MAT244Tests => Term Test 1 => Topic started by: Victor Ivrii on February 15, 2018, 05:07:40 PM

(a) Find the general solution for equation
\begin{equation*}
y''(t)3y'(t)+2y(t)=6+3 e^{t} .
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

$(a)$ $\\$
First we find the solution to the homogeneous solution for $$y’’(t)3y’(t)+2y(t)=0$$
Characteristic equation: $$r^23r+2$$
Hence, $$\cases{r_1=1\\r_2=2}$$ and the complementary solution for the homogeneous equation is $$y_c(t)=c_1e^t+c_2e^{2t}$$
Now we need to find the particular solution for the nonhomogeneous equation
$$y’’(t)3y’(t)+2y(t)=6+3e^t$$
Let $y_p(t)$ be the particular solution and $y_p(t)=Y_1(t)+Y_2(t)$.
We assume$Y_1(t)=At+B$, and there are no duplicates of the solution of the homogeneous equation, thus $Y_1’(t)=A$ and $Y_1’’(t)=0$. $\\$
And substitute these back to $y’’(t)3y’(t)+2y(t)=6$:
$$03A+2(At+B)=6 \implies \cases{A=0\\B=3}$$
We assume $Y_2(t)=Ce^t$, and noting there are duplicates of the solution of the homogeneous equation. $\\$ So we need to multiply $t$ for the propose of $Y_2(t)$, thus $Y_2(t)=Cte^t$ which leads to no duplicates. $\\$
Thus $Y_2’(t)=Ce^t+Cte^t$ and $Y_2’’(t)=2Ce^t+Cte^t$.
And substitute these values back to $y’’(t)3y’(t)+2y(t)=3e^t$:
$$2Ce^t+Cte^t3Ce^t3Cte^t+2Cte^t=3e^t \implies C=3$$
Hence, the particular solution $$y_p(t)=Y_1(t)+Y_2(t)=33e^t$$
Therefore, the general solution is $$y(t)=y_c(t)+y_p(t)=c_1e^t+c_2e^{2t}33te^t$$
$(b)$ $\\$
$$y’(t)=c_1e^t+2c_2e^{2t}3e^t3te^t$$
Set $t=0$ and $y=0$; $t=0$ and $y’=0$:
$$\cases{c_1+c_2=3\\c_1+2c_2=3} \implies \cases{c_1=3\\c_2=0}$$
Therefore, the solution for the IVP is $$y(t)=3e^t3te^t3$$