Series solution to a 2nd order linear ODE

Question

Consider the ODE $$(1+x^2)y''+3xy'+y=0$$ Find a solution in the form of $y=\sum a_nx^n$.

So after doing the algebra using these 2: $$y'=\sum_{n=1}^\infty n a_n(x-x_0)^{n-1}$$ $$y''=\sum_{n=2}^\infty n(n-1)a_n(x-x_0)^{n-2}$$ I got this relation: $$a_{n+2}=-a_n \cdot \frac{n+1}{n+2}$$ [Which is correct]. How do I get general solution? When n is odd i.e $n=1,2,3,\cdots$ I have $$a_3=-a_1\cdot \frac{2}{3} \\ a_5=-a_3\cdot \frac{4}{5} \\ a_7=-a_5\cdot \frac{6}{7}$$ and so on. The same is with the even $n$'s. I don't know how to write the general even $n$'s and the general odd $n$'s and finally the combination of the two.

Answer 1

Hint:

It seems that we have $$a_{n+2}=-a_n\cdot \frac{n+1}{n+2}$$ at least for every odd $n$. Perhaps, you should try to substitute the expression for $a_n$ in terms of $a_{n-2}$ into the above equation, and keep on repeating this step until you see an pattern that will lead you to your general expression.

Answer 2

Doing the same for the even powers of $x$, you find $$a_2= -\frac{a_0}{2}$$ $$a_4= -\frac{3 a_2}{4}$$ $$a_6 = -\frac{5 a_4}{6}$$ $$a_8= -\frac{7 a_6}{8}$$ Is this telling you something ?

Answer 3

HINT

I would say that

For even n: $a_{n+2}=-a_n \cdot \frac{n+1}{n+2}\Rightarrow y_1 = a_0(1-\frac{1}{2}x^2+\frac{1\cdot 3}{2\cdot 4}x^4-\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}x^6+\frac{1\cdot 3\cdot 5\cdot 7}{2\cdot 4\cdot 6\cdot 8}x^8- \cdots)$

The comparison with a series of $\frac{1}{\sqrt{1+x}}$ show, that $y_1=a_0\frac{1}{\sqrt{1+x^2}}$

The second solution $y_2$ is series for odd n $(a_1,a_3, a_5\cdots)$ - does not correspond to a known series. For the differential equation $y''+p(x)y'+q(x)y=0$ apply the formula

$\displaystyle y_2=y_1\int\frac{1}{y_1^2}e^{-\int p(x)dx} dx=\frac{1}{\sqrt{1+x^2}}\int (1+x^2)e^{-3\int \frac{x}{1+x^2}dx}dx=$

$=\displaystyle \frac{1}{\sqrt{1+x^2}}\int (1+x^2)e^{-\ln\sqrt{(1+x^2)^3}}dx=\frac{1}{\sqrt{1+x^2}}\int \frac{dx}{\sqrt{1+x^2}}=\frac{argsinh(x)}{\sqrt{1+x^2}}$

So I would say the solutin $\displaystyle y=a_0y_1+a_1y_2\Rightarrow y = C_1\frac{1}{\sqrt{1+x^2}}+C_2\frac{argsinh(x)}{\sqrt{1+x^2}}$