### Author Topic: TT1-problem 4  (Read 2517 times)

#### Victor Ivrii ##### TT1-problem 4
« on: October 09, 2014, 02:01:12 AM »
Find a particular solution of
\begin{equation*}
x^2 y''(x) - 6y(x)=10x^{-2} - 6, \qquad x >0 .
\end{equation*}

#### Yeming Wen

• Full Member
•   • Posts: 19
• Karma: 6 ##### Re: TT1-problem 4
« Reply #1 on: October 09, 2014, 09:28:21 AM »
Observe that the equation is an Euler equation, then we let
\begin{equation*} t = \log x
\end{equation*}
Then the equation becomes
\begin{equation*}
y'' - y' - 6y =10 e^{-2t}- 6
\end{equation*}
It is the same ODE from question 3 .
Use the particular solution from 3 i.e
\begin{equation*}
y=-2te^{-2t} + 1
\end{equation*}
Plug in $x$ back, we have
\begin{equation*} y=\frac{-2\log x}{x^2} + 1. \end{equation*}
« Last Edit: October 09, 2014, 10:06:16 AM by Victor Ivrii »

#### Victor Ivrii ##### Re: TT1-problem 4
« Reply #2 on: October 09, 2014, 10:11:42 AM »
I made some minor editing. Note that \log x, \sin ,\cos â. to keep them upright and provide a proper spacing.

But what would be a general solution? (I know it was not required in the test). Also I would like to see on the forum solution without substitution: i.e. we are looking at $y_p=y_{p1}+y_{p2}$, $y_{p1}=Ax^{-2}\ln x$ (because $r=-2$ is a characteristic root) and $y_{p2}=B$.
« Last Edit: October 21, 2019, 01:37:53 AM by Victor Ivrii »