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**Final Exam / Re: FE-2**

« **on:**April 18, 2013, 12:27:38 AM »

\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}-r-2 = 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}(t-1) + \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}

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}-r-2 = 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}(t-1) + \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}