Answer
$(Ly)(x)=8x^3\sin (2x^2+2)$
Work Step by Step
Given $L=4x^{2D}$
with $y_1=\sin^2(x^2+1)$
We first compute the appropriate derivatives
$y'(x)=\sin^2(x^2+1)(u'+u)=4x\sin (x^2+1)\cos (x^2+1)=2x\sin(2x^2+2)$
Hence
$(Ly)(x)=8x^3\sin (2x^2+2)$
