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TT1-problem 4

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**Victor Ivrii**:

Find a particular solution of

\begin{equation*}

x^2 y''(x) - 6y(x)=10x^{-2} - 6, \qquad x >0 .

\end{equation*}

**Yeming Wen**:

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*}

**Victor Ivrii**:

I made some minor editing. Note that \log x, \sin ,\cos รข. 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 $y_{p2}=B$.

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