The differential equation representing family of curves \(y = e^{-2x}(a\cos x + b\sin x)\), where a and b are arbitrary constants, is :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\displaystyle\frac{d^{2}y}{dx^{2}}+4\frac{dy}{dx}+5y=0\)
Two arbitrary constants are to be eliminated, so the curve is differentiated twice. Written as
\(y\,e^{2x} = a\cos x + b\sin x\)
Differentiating once:
\(e^{2x}\left(y' + 2y\right) = -a\sin x + b\cos x\)
Differentiating again:
\(e^{2x}\left(y'' + 2y'\right) + 2e^{2x}\left(y' + 2y\right) = -a\cos x - b\sin x\)
The right side is \(-y\,e^{2x}\). Dividing by \(e^{2x}\):
\(y'' + 2y' + 2y' + 4y = -y\)
\(y'' + 4y' + 5y = 0\)
That is, \(\displaystyle\frac{d^{2}y}{dx^{2}}+4\frac{dy}{dx}+5y=0\).