If $y =\sin \log (x^2+2x+1)$ then $(x+1)^2 y^{(n+2)}+(2n+1)(x+1)y^{(n+1)}+(x^2+4)y^{(n)}=0$ How to prove this?


  If $y =\sin \log (x^2+2x+1)$, and $y'=y^{(1)}$, $y''=y^{(2)}$, and so on, prove that $(x+1)^2 y^{(n+2)}+(2n+1)(x+1)y^{(n+1)}+(x^2+4)y^{(n)}=0 $


Finding $y'$ is simple, but $y''$ and other higher terms seems difficult. Any short solution for this approach?
$y=\sin 2\log(x+1)
=2\sin\log(x+1)\cos\log(x+1)$.

Apply Leibnitz's formula for the n-th derivative of a product and look term-by-term.
$y=\sin 2\log(x+1)
=2\sin\log(x+1)\cos\log(x+1)$.

Apply Leibnitz's formula for the n-th derivative of a product and look term-by-term.