Toronto Math Forum
MAT2442013S => MAT244 MathTests => Final Exam => Topic started by: Victor Ivrii on April 17, 2013, 03:01:31 PM

Find the general solution of the equation
\begin{equation*}
t^2y''2y =t^4 e^t,\qquad t>0.
\end{equation*}

Heres a solution

Matthew, it is unreadable

\begin{equation}
t^{2}y"  2y = t^{4}e^{t}, t>0
\end{equation}
Solve Euler for homogeneous, and find the two solutions to the system:
\begin{equation}
\begin{split}
x = \ln(t)\\
y" y' 2y = 0\\
r^{2}r2 = 0\\
r_1 = 1, r_2 = 2\\
\end{split}
\end{equation}
\begin{equation}
\begin{split}
y_1 = e^{x} = t^{1}\\
y_2 = e^{2x} = t^2
\end{split}
\end{equation}
Then, using the formula for variation of parameters, you get:
\begin{equation}
y_p(t) = \frac {t^{2}}{3} \int{te^{t}dt} + \frac{t^{1}}{3} \int{t^{4}e^{t}dt}
\end{equation}
\begin{equation}
y_p(t) = \frac{t^{2}e^t}{3}(t1) + \frac{t^{1}e^{t}}{3}(t^{4}  4t^{3} + 12t^{2}  24t + 24)
\end{equation}
Thus, the general solution is given by:
\begin{equation}
Y(t) = c_{1}t^{1} + c_2t^{2} + y_{p}(t)
\end{equation}

Sorry about that upload I attatched a better quality one

Now its OK. Naveed, you don't need to plug $x=\ln(t)$, just wright down $r(r1)2=0$ directly.