If \(y=a \sin x+b \cos x\) (where a and b are constants), then \(y^2+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2\) is

\(y=\mathrm{a} \sin x+\mathrm{b} \cos x\)
Differentiating w.r.t. \(x\), we get
\(\frac{\mathrm{d} y}{\mathrm{~d} x}=\mathrm{a} \cos x-\mathrm{b} \sin x\)
\(\therefore \quad\) Now,
\(\begin{aligned}
& y^2+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2 \\
= & (\mathrm{a} \sin x+\mathrm{b} \cos x)^2+(\mathrm{a} \cos x-\mathrm{b} \sin x)^2 \\
= & \mathrm{a}^2 \sin ^2 x+\mathrm{b}^2 \cos ^2 x+2 \mathrm{ab} \sin x \cos x+\mathrm{a}^2 \cos ^2 x \\
& +\mathrm{b}^2 \sin ^2 x-2 \mathrm{ab} \sin x \cdot \cos x \\
= & \mathrm{a}^2\left(\sin ^2 x+\cos ^2 x\right)+\mathrm{b}^2\left(\sin ^2 x+\cos ^2 x\right) \\
= & \mathrm{a}^2+\mathrm{b}^2 \\
\therefore \quad & y^2+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2=\mathrm{a}^2+\mathrm{b}^2
\end{aligned}\)
Here, \(\mathrm{a}, \mathrm{b}\) are constants.
\(\therefore \quad y^2+\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)^2\) is also a constant.