- #1
Caglar Yildiz
- 19
- 0
Hi
I am supposed to find solution of $$xy''+y'+xy=0$$
but i am left with reversing this equation.
i am studying solution of a differential equation by series now and I cannot reverse a series in the form of:
$$ J(x)=1-\frac{1}{x^2} +\frac{3x^4}{32} - \frac{5x^6}{576} ...$$
$$ \frac{1}{J}=1+\frac{x^2}{2} +\frac{5x^4}{32}+ \frac{23x^6}{576}...$$
General formula of $J(x)$ is $$\sum_{n=0}^{\infty} \frac{(-1)n}{(n!)^2}(\frac{x}{2})^2$$
Thanks for all help!
I am supposed to find solution of $$xy''+y'+xy=0$$
but i am left with reversing this equation.
i am studying solution of a differential equation by series now and I cannot reverse a series in the form of:
$$ J(x)=1-\frac{1}{x^2} +\frac{3x^4}{32} - \frac{5x^6}{576} ...$$
$$ \frac{1}{J}=1+\frac{x^2}{2} +\frac{5x^4}{32}+ \frac{23x^6}{576}...$$
General formula of $J(x)$ is $$\sum_{n=0}^{\infty} \frac{(-1)n}{(n!)^2}(\frac{x}{2})^2$$
Thanks for all help!