Differential Equations Question 125
Question: Family of curves $ y=e^{x}(A\cos x+B\sin x) $ , represents the differential equation
[MP PET 1999]
Options:
A) $ \frac{d^{2}y}{dx^{2}}=2\frac{dy}{dx}-y $
B) $ \frac{d^{2}y}{dx^{2}}=2\frac{dy}{dx}-2y $
C) $ \frac{d^{2}y}{dx^{2}}=\frac{dy}{dx}-2y $
D) $ \frac{d^{2}y}{dx^{2}}=2\frac{dy}{dx}+y $
Show Answer
Answer:
Correct Answer: B
Solution:
Given $ y=e^{x}A\cos x+e^{x}B\sin x $
$ \frac{dy}{dx}=Ae^{x}\cos x-Ae^{x}\sin x+Be^{x}\sin x+Be^{x}\cos x $
$ \frac{dy}{dx}=(A+B)e^{x}\cos x+(B-A)e^{x}\sin x $
$ \frac{d^{2}y}{dx^{2}}=(A+B)e^{x}\cos x-e^{x}\sin x(A+B) $
$ +(B-A)e^{x}\sin x+(B-A)e^{x}\cos x $
$ \frac{d^{2}y}{dx^{2}}=2Be^{x}\cos x-2Ae^{x}\sin x $ . Hence $ \frac{d^{2}y}{dx^{2}}=2\frac{dy}{dx}-2y $ .
BETA
