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

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

we had to differentiate given expression twice to eliminate A and B from equation

$y=e^{x}(A \cos x+B\sin x)...eq1 \\ \\ y = e^{x}(A \cos x+ B\sin x )\\ \\ \frac{dy}{dx} = {e}^{x}( - A\: sin \: x + B \: cos \: x) + e^{x}(A \cos x+B\sin x) \\ \\ from \: eq1 \\ \\ \frac{dy}{dx} = {e}^{x}( - A \: sin \: x + B \: cos \: x) +y ...eq2\\ \\ differentiate \: again \\ \\ \frac{ {d}^{2}y }{ {dx}^{2} } = {e}^{x} ( - A \: cos \: x - B\: sin \: x) + {e}^{x}( - A \: sin \: x + B \: cos \: x) + \frac{dy}{dx} \\ \\ \frac{ {d}^{2}y }{ {dx}^{2} } = - {e}^{x} ( A \: cos \: x + B \: sin \: x) + {e}^{x}( - A \: sin \: x + B \: cos \: x) + \frac{dy}{dx} \\ \\ \frac{ {d}^{2}y }{ {dx}^{2} } = - y + \frac{dy}{dx} - y + \frac{dy}{dx} \: \: \: from \: eq1 \: and \: eq2 \\ \\ \frac{ {d}^{2}y }{ {dx}^{2} } = - 2y + 2 \frac{dy}{dx} \\ \\ \frac{ {d}^{2}y }{ {dx}^{2} } - 2 \frac{dy}{dx} + 2y = 0$
is the final solution.