Question

Find the differential equation of family of curves $y=e^x(A\cos x+B\sin x)$ where $A$ and B are arbitrary constants.

Answer 1

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

$y_1=e^x(-A\sin x+B\cos x)+e^x(A\cos x+B\sin x)$

$y_1=e^x\left(A\right(\cos x-\sin x)+B(\cos x+\sin x\left)\right)$

$y_2=e^x\left(A\right(-\sin x-\cos x)+B(\sin x+\cos x\left)\right)+e^x\left(A\right(\cos x-\sin x)+B(\cos x+\sin x\left)\right)$

$y_2=e^x\left(A\right(-2\sin x\left)\right)+B\left(2\cos x\right))$

$y_2=2e^x(B\cos x-A\sin x)$

As, $y+\frac{y_2}{2}=y_1$

$y_2-2y_1+2y=0$