Browse Questions

# True-or-False: Differential equation representing the family of curves $y=e^x(A\cos x+B\sin x)$ is $\frac{d^2y}{dx^2} - 2\frac{dy}{dx} + 2y = 0$.

$\begin{array}{1 1} True \\ False \\ Cannot\;be\;determined \end{array}$

Toolbox:
• If the given equation has $n$ arbitrary constants then its differential eqaution will be of $n^{th}$ order.
• Product rule: $\large\frac{d}{dx}$$(uv)=u. \large\frac{dv}{dx}$$+v.\large\frac{du}{dx}$
• $\int e^{-ax}dx=-\large\frac{1}{a}$$e^{ax} y=e^x(A \cos x+B \sin x) This can be written as y e^{-x}=A\cos x+B \sin x ye^{-x} is a product of two functions, Hence it can be differentiated by using product rule \large\frac{d}{dx}$$(uv)=u. \large\frac{dv}{dx}$$+v.\large\frac{du}{dx} Let u=y, then \large\frac{du}{dx}=\frac{dy}{dx} and let v=e^{-x} then \large\frac{dv}{dx}=-e^{-x} \large\frac{d}{dx}$$(uv)=-y.e^{-x}+e^{-x}.\large\frac{dy}{dx}$
Hence differentiating $ye^{-x}=A \cos x+B \sin x$