### Author Topic: FE-2  (Read 4153 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*}
\end{equation*}

#### Matthew Cristoferi-Paolucci

• Jr. Member
• Posts: 10
• Karma: 8
##### 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 »

#### Hareem Naveed

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##### Re: FE-2
« Reply #3 on: April 18, 2013, 12:27:38 AM »

t^{2}y" - 2y = t^{4}e^{t},    t>0

Solve Euler for homogeneous, and find the two solutions to the system:

\begin{split}
x = \ln(t)\\
y"- y'- 2y = 0\\
r^{2}-r-2 = 0\\
r_1 = -1,  r_2 = 2\\
\end{split}

\begin{split}
y_1 = e^{-x} = t^{-1}\\
y_2 = e^{2x} = t^2
\end{split}

Then, using the formula for variation of parameters, you get:

y_p(t) = -\frac {t^{2}}{3} \int{te^{t}dt} + \frac{t^{-1}}{3} \int{t^{4}e^{t}dt}

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)

Thus, the general solution is given by:

Y(t) = c_{1}t^{-1} + c_2t^{2} + y_{p}(t)

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

#### Matthew Cristoferi-Paolucci

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• Posts: 10
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##### Re: FE-2
« Reply #4 on: April 18, 2013, 12:38:05 AM »
Now its OK. Naveed, you don't need to plug $x=\ln(t)$, just wright down $r(r-1)-2=0$ directly.