Q. If $y = e^{ax}\sin{bx,\ \ then\frac{d^{2}y}{dx^{2}} - 2a\frac{dy}{dx} + a^{2}y}\$is equal to

A) 0
B) 1
C) $- b^{2}y$
D) $- by$
Explanation

(c)
Given, $y = e^{ax}\sin{bx}$ …(i)
$\Rightarrow \ \ \ \ \ \ \ \ \ \frac{dy}{dx} = ae^{ax}\sin{bx + be^{ax}\cos{bx}}$
$\Rightarrow \ \ \ \ \ \frac{dy}{dx}\ = \ ay + be^{ax}\cos{bx}\$ …(ii)
$\Rightarrow \ \ \frac{d^{2}y}{dx^{2}} = a\frac{dy}{dx} + abe^{ax}\cos{bx - e^{ax}b^{2}\sin{bx}}$
$\Rightarrow \ \ \frac{d^{2}y}{dx^{2}} = a\frac{dy}{dx} + a\left( \frac{dy}{dx} - ay \right) - b^{2}y$
[from Eqs. (i) and (ii)]
$\Rightarrow \ \ \frac{d^{2}y}{dx^{2}} - 2a\frac{dy}{dx} + a^{2}y = - b^{2}y$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.