Question

If $y=a\sin x+b\cos x$, then $y^2+{\left(\frac{dy}{dx}\right)}^2$ is

• A. function of $x$
• B. function of $y$
• C. function of $x$ and $y$
• D. constant

Answer 1

The correct option is D constant

$y=a\sin x+b\cos x$

$\frac{dy}{dx}=a\cos x-b\sin x$

$y^2+{\left(\frac{dy}{dx}\right)}^2={(a\sin x+b\cos x)}^2+{(a\cos x-b\sin x)}^2$

$=a^2{\sin }^{2}x+b^2{\cos }^{2}x+2ab\sin x\cos x+a^2{\cos }^{2}x+b^2{\sin }^{2}x-2ab\sin x\cos x$

$=a^2({\sin }^{2}x+{\cos }^{2}x)+b^2({\sin }^{2}x+{\cos }^{2}x)$

$=a^2+b^2$ since ${\sin }^{2}x+{\cos }^{2}x=1$

$=constant$