Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:29:15 AM

(a) Find the general solution for equation
\begin{equation*}
y''5y'+4y=12e^t + 20 e^{t}.
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

I will type the solution soon!

Since the first post did not type out the solution, I will type mine here.
Thank you Shengying, I fixed my problem here.
First, we need to find the homogeneous solution
$r^25r+4=0$
$(r4)(r1)=0$
$r_1=4,r_2=1$
$y_c(t) = c_1e^{4t}+c_2e^{t}$
Secondly, we need to find particular solution of $y"  5y' +4y = 12e^t$
$y_p(t) =Ate^t, y'_p(t) = Ae^t +Ate^t, y"_p(t) = 2Ae^t+Ate^t$
plug in back to the equation we get A=4
Thus, $y_p(t) =4te^t$
Thirdly, we need to find particular solution of $y"  5y' +4y = 20e^{t}$
$y_p(t) =Bte^{t}, y'_p(t) = Be^{t}, y"_p(t) = Be^{t}$
plug in back to the equation we get B=2
Thus, $y_p(t) =2e^{t}$
Therefore, $y(t) = c_1e^4t +c_2e^t+4te^t+2e^{t}$
(b) $y(0) = 0$ so that $c_1+c_2 = 2$
$y'(0) =0$ so that $4c_1+c_2 = 2$
Therefore $c_1=0, c_2 = 2$ and $y(t) = 2e^t +4te^t+2e^{t}$

Zixuan, I think you made a mistake.
Plugging in $y(0)=0$ , you should get $0=C_1+C_2+2=0$. Therefore, $C_1=1, C_2=2$
$$∴y=2e^t+4te^t+2e^{t}$$

Everybody was right: Zhihao solved correctly, Zixua typed (you need to learn a better way to type it) and Shengying found a mistype in the solution (not the answer)