Consider the initial value problem $$xu''+\sin(x)u=0 \ \ \ \ u(0)=0, u'(0)=2$$ Derive the first $4$ non-zero terms of a power series solution to this problem about the point $x=0$.
I know the solution will have the from $$u(x)=\sum_{k=0}^{\infty}A_kx^k$$ But upon differentiating, I get that $$\sum_{k=0}^{\infty}k(k-1)A_kx^{k-2}+\sin(x)\sum_{k=0}^{\infty}A_kx^{k-1}=0$$ and do not know how to proceed. Would an initial substitution help?
I willshow you how to proceed following daruma's comment. Let $$ u=2\,x +A_2\,x^2+A_3\,x^3+\dots $$ Then $$ x\,u''=2\,A_2\,x+6\,A_3\,x^2+\dots $$ and \begin{align} (\sin x)u&=\Bigl(x-\frac{x^3}{6}+\dots\Bigr)\bigl(2\,x +A_2\,x^2+A_3\,x^3+\dots\bigr)\\ &=2\,x^2+A_2\,x^3+\Bigl(A_3-\frac13\Bigr)\,x^4+\dots \end{align} Now sum the series for $x\,u''$ and $(\sin x)u$, set it equal to $0$ and obtain equations for the coefficients $A_n$, $n\ge2$.