(D)
We have, $x=a \sin t-b \cos t$ and $y=a \cos t+b \sin t$
$\begin{array}{l}
x^{2}+y^{2}=\left\{[a \sin t-b \cos t]^{2}+[a \cos t+b \sin t]^{2}\right\} \\
x^{2}+y^{2}=\left\{a^{2} \cdot \sin ^{2} t-2 a b \sin t \cdot \cos t+b^{2} \cos ^{2} t+a^{2} \cos ^{2} t+2 a b \sin t \cdot \cos t+b^{2} \sin ^{2} t\right\} \\
x^{2}+y^{2}=\left\{a^{2}\left(\sin ^{2} t+\cos ^{2} t\right)+b^{2}\left(\cos ^{2} t+\sin ^{2} t\right)\right\} \\
x^{2}+y^{2}=a^{2}+b^{2}
\end{array}$
Differentiate w.r.t. $x$, we get
$2 x+2 y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{x}{y}$...(1)
$\therefore \quad \mathrm{y} \frac{\mathrm{dy}}{\mathrm{dx}}=-\mathrm{x}$
Again differentiating w.r.t. x, we get
$y \frac{d^{2} y}{d x^{2}}+\left(\frac{d y}{d x}\right)^{2}=-1$
Substituting value of $\frac{d y}{d x}$ from (1), we get
$\begin{aligned}
& y \frac{d^{2} y}{d x^{2}}+\left(\frac{-x}{y}\right)^{2}=-1 \Rightarrow y \frac{d^{2} y}{d x^{2}}+\frac{x^{2}}{y^{2}}=-1 \\
\therefore & y^{3} \frac{d^{2} y}{d x^{2}}+x^{2}+y^{2}=0
\end{aligned}$