What is the power series, (or summation form) for the following equation? I know the first couple of terms, but am unable to write it as a power series. The equation is dy/dx +x*y = x^2 There is also the condition y(0)=0
So far I have managed to work out that the third term is 1/3 the fifth term is -1/15 and so on. The other terms are all 0.
$\frac{dy}{dx}+xy=x^2$. Let $g=e^{x^2/2}$ then $g(0)=1$, $g'(0)=0$.
$\frac{d}{dx}(g y)=x^2g$ by integrating factor.
$g'y+gy'=x^2g \implies y'(0)=0.$
By Frobenius' Method.
$xy=\sum_{k=0}^\infty c_kx^{p+k+1}$
$y'=\sum_{k=0}^\infty c_k(p+k)x^{p+k-1}=c_0px^{p-1}+c_1(p+1)x^p+\sum_{k=0}^\infty c_{k+2}(p+k+2)x^{p+k+1}$
So:
$y'+xy=c_0px^{p-1}+c_1(p+1)x^p+\sum_{k=0}^\infty [c_{k+2}(p+k+2)+c_k]x^{p+k+1}=x^2$
So we need $\frac{c_{k+2}}{c_k}=\frac{-1}{(p+k+2)}$
$c_0px^{p-3}+c_1(p+1)x^{p-2}+\sum_{k=0}^\infty [c_{k+2}(p+k+2)+c_k]x^{p+k-1}=1$
Suppose $c_0=0$. If we only allow non-negative powers of $x$, then $p=2$ and $c_1=1/3$.
If $c_1=0$ then $p=3$ and $c_0=1/3$. In either case, the iterative process above generates the other coefficients defining the function.
Case $c_1=0$: $c_0=1/3, c_2=-1/15,c_4=1/105,...$
Case $c_0=0$: $c_1=1/3, c_3=-1/15, c_5=1/105, ...$