Toronto Math Forum
MAT2442018F => MAT244Tests => Term Test 1 => Topic started by: Victor Ivrii on October 16, 2018, 05:35:54 AM

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

answer

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
$r^25r8=0$
$(r+4)(r2)=0$
$r_1=4,r_2=2$
$y_c(t) = c_1e^{4t}+c_2e^{2t}$
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 $y(t) = \frac{7}{6}e^{4t} +\frac{11}{6}e^{2t}5te^{2t}3e^{2t}$