Answer to Question #162083 in Differential Equations for Gul Rehman

Question

x^2 y^n-2x(1+x)y'+2(1+x)y=x^3

Answer 1

$y(x)=\displaystyle\sum_{k=0}^{\infin}a_kx^k$ , $y'(x)=\displaystyle\sum_{k=1}^{\infin}ka_kx^{k-1}$ , $y^{(n)}(x)=\displaystyle\sum_{k=n}^{\infin}\frac{k!}{(k-n)!}a_kx^{k-n}$

$x^2\displaystyle\sum_{k=n}^{\infin}\frac{k!}{(k-n)!}a_kx^{k-n}-2x(x+1)\displaystyle\sum_{k=1}^{\infin}ka_kx^{k-1}+2(x+1)\displaystyle\sum_{k=0}^{\infin}a_kx^k=x^3$

$\displaystyle\sum_{k=n}^{\infin}\frac{k!}{(k-n)!}a_kx^{k-n+2}-2\displaystyle\sum_{k=1}^{\infin}ka_kx^{k+1}-2\displaystyle\sum_{k=1}^{\infin}ka_kx^{k}+$

$+2\displaystyle\sum_{k=0}^{\infin}a_kx^{k+1}+2\displaystyle\sum_{k=0}^{\infin}a_kx^k=x^3$

$\displaystyle\sum_{k=2}^{\infin}\frac{(k+n-2)!}{(k-2)!}a_{k+n-2}x^{k}-2\displaystyle\sum_{k=2}^{\infin}(k-1)a_{k-1}x^{k}-2\displaystyle\sum_{k=1}^{\infin}ka_kx^{k}+$

$+2\displaystyle\sum_{k=1}^{\infin}a_{k-1}x^{k}+2\displaystyle\sum_{k=0}^{\infin}a_kx^k=x^3$

$k=0\implies 2a_0=0\implies a_0=0$

$k=2\implies n!a_n-2a_1-4a_2+2a_1+2a_2=0$

$a_n=\frac{2a_2}{n!}\implies a_2=n!a_n/2$

$k=3\implies (n+1)!a_{n+1}-4a_2-6a_3+2a_2+2a_3=1$

$a_3=\frac{(n+1)!a_{n+1}-n!a_n-1}{4}$

$k>3\implies \frac{(k+n-2)!}{(k-2)!}a_{k+n-2}+2a_{k-1}(2-k)+2a_k(1-k)=0$

$a_k=\frac{\frac{(k+n-2)!}{(k-2)!}a_{k+n-2}+2a_{k-1}(2-k)}{2(k-1)}$

$y(x)=a_1x+\frac{a_nn!}{2}x^2+\frac{(n+1)!a_{n+1}-n!a_n-1}{4}x^3+...+\frac{\frac{(k+n-2)!}{(k-2)!}a_{k+n-2}+2a_{k-1}(2-k)}{2(k-1)}x^k$