MAT244--2018F > Term Test 1

TT1 Problem 3 (afternoon)

(1/1)

Victor Ivrii:
(a) Find the general solution for equation
\begin{equation*}
y''+2y'-8y=-30e^{2t} + 24 e^{-2t}.
\end{equation*}
(b) Find solution, satisfying $y(0)=0$, $y'(0)=0$.

Xiaoyuan Wang:

Zixuan Wang:
Since the first post did not type out the solution, I will type out my solution and fix my mistake in the picture.
First, we need to find the homogeneous solution
$r^2-5r-8=0$
$(r+4)(r-2)=0$
$r_1=-4,r_2=2$
$y_c(t) = c_1e^{-4t}+c_2e^{2t}$
Secondly, we need to find particular solution of $y" +2y' -8y = -30e^{2t}$
$y_p(t) =Ate^{2t}, y'_p(t) = Ae^{2t} +2Ate^{2t}, y"_p(t) = 4Ae^{2t}+4Ate^t$
$4Ae^{2t}+4Ate^{2t}+2Ae^{2t}+4Ate^{2t}-8Ate^{2t}=-30e^{2t}$
$6Ae^{2t} = -30e^{2t}$
A=-5
Thus, $y_p(t) =-5te^{2t}$
Thirdly, we need to find particular solution of $y" +2y' -8y = 24e^{-2t}$
$y_p(t) =Bte^{-2t}, y'_p(t) = -2Be^{-2t}, y"_p(t) = 4Be^{-2t}$
plug in back to the equation we get,
$4Be^{-2t}-4Be^{-2t} - 8Be^{-2t}=24e^{2t}$
Thus, B=-3
Thus, $y_p(t) =-3e^{-2t}$
Therefore, $y(t) = c_1e^{-4t} +c_2e^{2t}-5te^{2t}-3e^{-2t}$
(b) $y(0) = 0$ so that $c_1+c_2 -3= 0$
$y'(0) =0$ so that $-4c_1+2c_2 +1= 0$
Therefore $c_1=\frac{7}{6}, c_2 = \frac{11}{6}$ and $y(t) = \frac{7}{6}e^{-4t} +\frac{11}{6}e^{2t}-5te^{2t}-3e^{-2t}$