Family of curves $y=e^x(A\,cosx+B\,sinx)$, represents the differential equation:
Answer & explanation
Correct answer: option 2
$y=e^x(A\cos x+B\sin x)$
$y'=\frac{dy}{dx}=e^x(-A\sin x+B\cos x)+e^x(A\cos x+B\sin x)$
$y'=A(\cos x-\sin x).e^x+B(\sin x+\cos x)e^x$
$y''=A\{[-\sin x-\cos x]e^x+e^x[\cos x-\sin x]\}+B\{[\cos x-\sin x]e^x+[\sin x+\cos x]e^x\}$
$⇒y''=A(-2\sin x)e^x+B(2\cos x)e^x$
$y''=2y'-2y$ or $\frac{d^2y}{dx^2}=2\frac{dy}{dx}-2y$