Author Topic: FE-2  (Read 1906 times)

Victor Ivrii

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FE-2
« 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*}

Matthew Cristoferi-Paolucci

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Re: FE-2
« Reply #1 on: April 17, 2013, 05:42:42 PM »
Heres a solution

Victor Ivrii

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Re: FE-2
« Reply #2 on: April 17, 2013, 11:16:47 PM »
Matthew, it is unreadable

Hareem Naveed

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Re: FE-2
« Reply #3 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}

« Last Edit: April 18, 2013, 01:04:05 AM by Victor Ivrii »

Matthew Cristoferi-Paolucci

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Re: FE-2
« Reply #4 on: April 18, 2013, 12:38:05 AM »
Sorry about that upload I attatched a better quality one

Victor Ivrii

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Re: FE-2
« Reply #5 on: April 18, 2013, 01:06:30 AM »
Now its OK. Naveed, you don't need to plug $x=\ln(t)$, just wright down $r(r-1)-2=0$ directly.