Question

Find the differential equation of the family of curve $y=e^x(A\cos x+B\sin x)$.

Answer 1

given the family of curve is $y=e^x(A\cos x+B\sin x)...\left(1\right)$

Differentiating on both sides

$=A(e^x\cos x-e^x\sin x)+B(e^x\sin x+e^x\cos x)....\left(2\right)$

Again differentiating

$=A(e^x\cos x-e^x\sin x-e^x\sin x-e^x\cos x)+B(e^x\sin x+e^x\cos x+e^x\cos x-e^x\sin x)$

$=-2A{e}^x\sin x+2B{e}^x\cos x....\left(3\right)$

from 2

y'$=e^x(A\cos x+B\sin x)+e^x(-a\sin x+b\cos x)$

y'$=y+\frac{y^{\prime \prime }}{2}$