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Find the general solution of $$(1+x^2)y''+2xy'-2y=0$$

in terms of power series in $x$. Can you express this solution by means of elementary functions?

I know that $y= \displaystyle\sum_{n=0}^{ \infty } a_nx^n$ and $y'= \displaystyle\sum_{n=1}^{ \infty } a_nnx^{n-1}$ and $y''=\displaystyle\sum_{n=2}^{ \infty } a_nn(n-1)x^{n-2}$.

As far as I can tell I am to simply plug these summations in the original equation stated above. Also, I shifted the following to get $x^n$. Giving:

$$\sum_{n=0}^{ \infty } a_{n+2}(n+2)(n+1)x^{n}+\sum_{n=2}^{ \infty } a_nn(n-1)x^{n}+\sum_{n=1}^{ \infty }2 a_nnx^{n}-\sum_{n=0}^{ \infty } 2a_nx^n=0$$

Now, I combine the equation into a single summation. $$\sum_{n=2}^{ \infty }\bigg[a_{n+2}(n+2)(n+1)+a_nn(n-1)+2a_nn-2a_n\bigg]x^n=0$$

Doing this, I am left over with $2(a_2-a_0)=0$ and $6a_3x=0$. I can also form $a_{n+2}=a_n\frac{1-n}{n+1}$ Following this I plugged values in for the third equation, but this is where things start to get difficult. I'm honestly not sure how to continue. The question itself confuses me specifically "in terms of power series in $x$. Can you express this solution by means of elementary functions?".

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1 Answer 1

Let $y=\sum\limits_{n=0}^\infty a_nx^n$ ,

Then $y'=\sum\limits_{n=0}^\infty na_nx^{n-1}$

$y''=\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}$

$\therefore(1+x^2)\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}+2x\sum\limits_{n=0}^\infty na_nx^{n-1}-2\sum\limits_{n=0}^\infty a_nx^n=0$

$\sum\limits_{n=0}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=0}^\infty n(n-1)a_nx^n+\sum\limits_{n=0}^\infty2na_nx^n-\sum\limits_{n=0}^\infty2a_nx^n=0$

$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=0}^\infty(n^2+n-2)a_nx^n=0$

$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=0}^\infty(n+2)(n-1)a_nx^n=0$

$\sum\limits_{n=2}^\infty n(n-1)a_nx^{n-2}+\sum\limits_{n=2}^\infty n(n-3)a_{n-2}x^{n-2}=0$

$\sum\limits_{n=2}^\infty(n(n-1)a_n+n(n-3)a_{n-2})x^{n-2}=0$

$\therefore n(n-1)a_n+n(n-3)a_{n-2}=0$

$a_n=-\dfrac{(n-3)a_{n-2}}{n-1}$

$\therefore\begin{cases}a_1=a_1\\a_{2n+3}=0~\forall n\in\mathbb{Z}^*\\a_0=a_0\\a_{2n}=\dfrac{(-1)^n((-1)1\times3\times......(2n-3))a_0}{1\times3\times5\times......(2n-1)}\forall n\in\mathbb{N}\end{cases}$

$\begin{cases}a_1=a_1\\a_{2n+3}=0~\forall n\in\mathbb{Z}^*\\a_0=a_0\\a_{2n}=\dfrac{(-1)^{n+1}a_0}{2n-1}\forall n\in\mathbb{N}\end{cases}$

$\begin{cases}a_1=a_1\\a_{2n+3}=0~\forall n\in\mathbb{Z}^*\\a_{2n}=\dfrac{(-1)^{n+1}a_0}{2n-1}\forall n\in\mathbb{Z}^*\end{cases}$

$\therefore y=C_1x+C_2\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+1}x^{2n}}{2n-1}=C_1x+C_2\biggl(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^{n+1}x^{2n}}{2n-1}\biggr)=C_1x+C_2\biggl(1+\sum\limits_{n=0}^\infty\dfrac{(-1)^nx^{2n+2}}{2n+1}\biggr)=C_1x+C_2(1+x\tan^{-1}x)$

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