Find a solution of $$ u_{tt}-u_{xx}=\lambda^2u$$ of a form $$u=f(x^2-t^2)=f(s) \text{ with } f(0)=1,$$ in form of power series in s, where $\lambda$ is a nonzero constant.
First thing I do is to find $u_{tt}$ and $u_{xx}$ $$u_{tt}=f''(s)4t^2-2f'(s)$$ $$u_{xx}=f''(s)4x^2+2f'(s)$$ Thus, we have $$u_{tt}-u_{xx}=f''(s)4(t^2-x^2)-4f'(s)=\lambda^2u$$ That is $$f''(s)4s+4f'(s)+\lambda^2f(s)=0$$ If we assume $$f(s)=\sum_{k=0}^{\infty}A_k s^k,$$ then $$4s\sum_{k=2}^\infty k(k-1)A_ks^{k-2}+4\sum_{k=1}^\infty k A_k s^{k-1}+\lambda^2\sum_{k=0}^\infty A_k s^k=0.$$
Also, we can written as $$4\sum_{k=2}^\infty k(k-1)A_ks^{k-1}+4\sum_{k=1}^\infty k A_k s^{k-1}+\lambda^2\sum_{k=0}^\infty A_k s^k=0.$$ Also equal to $$4\sum_{k=2}^\infty k(k-1)A_ks^{k-1}+4\sum_{k=1}^\infty k A_k s^{k-1}+\lambda^2\sum_{k=1}^\infty A_{k-1} s^{k-1}=0.$$
$$\sum_{k=2}^\infty [4k(k-1)A_k+4k A_k+\lambda^2 A_{k-1}]s^{k-1}+4A_1+\lambda^2A_0=0.$$
Compare the coefficient for $s^0$: $$4A_1+\lambda^2A_0=0$$ Since $A_0=1$ form the conditon $f(0)=1,$ we have $A_1=\frac{-\lambda^2}{4}.$
Compare the coefficient for $s^k(k\geq 1)$: $$4k(k-1)A_k+4k A_k +\lambda^2 A_{k-1} =4k^2A_k+\lambda^2 A_{k-1}=0$$ Therefore, $A_k=\frac{-\lambda^2}{4k^2}A_{k-1}$ for $k\geq 2.$
Thus, $u=f(s)=\sum_{k=0}^{\infty}A_k s^k,$ with $A_0=1$ and $A_k=\frac{-\lambda^2}{4k^2}A_{k-1}$ for $k\geq 1.$
Is my answer right?
If right, how can I check my answer?