$\begin{align}  & y={{x}^{2}}\sin x \\ & \frac{dy}{dx}={{x}^{2}}\tfrac{d}{dx}(\sin x)+\sin x\tfrac{d}{dx}({{x}^{2}}) \\ & \frac{dy}{dx}={{x}^{2}}\cos x+2x\sin x \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=-{{x}^{2}}\sin x+2x\cos x+2x\cos x+2\sin x \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=-{{x}^{2}}\sin x+4x\cos x+2\sin x \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=4x\cos x+2\sin x-y \\ & \text{Multiply through by }{{x}^{2}} \\ & {{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=4{{x}^{3}}\cos x+2x\sin x-{{x}^{2}}y \\ & \text{Add }(8{{x}^{2}}\sin x)\text{ and subtract }8{{x}^{2}}\sin x\text{ to R}\text{.H}\text{.S} \\ & {{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=4{{x}^{3}}\cos x+8{{x}^{2}}\sin x-8{{x}^{2}}\sin x+2{{x}^{2}}\sin x-{{x}^{2}}y \\ & {{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=4x(\underbrace{{{x}^{2}}\cos x+2x\sin x}_{\tfrac{dy}{dx}})-6\underbrace{{{x}^{2}}\sin x}_{y}-{{x}^{2}}y \\ & {{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=4x\frac{dy}{dx}-6y-{{x}^{2}}y \\ & {{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}-4x\frac{dy}{dx}+6y+{{x}^{2}}y=0 \\ & {{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}-4x\frac{dy}{dx}+(6+{{x}^{2}})y=0 \\ & \text{Note }y={{x}^{2}}\sin x,\text{ }\frac{dy}{dx}={{x}^{2}}\cos x+2x\sin x \\\end{align}$