Q. The general solution of the differential equation $\frac{d^{2}y}{dx^{2}} + 2\frac{dy}{dx} + y = 2e^{3x}$ is given by
A) $y = \left( c_{1} + c_{2}x \right)\ e^{x} + \frac{e^{3x}}{8}\$
B) $y = \left( c_{1} + c_{2}x \right)e^{- x} + \frac{e^{- 3x}}{8}$
C) $y = \left( c_{1} + c_{2}x \right)e^{- x} + \frac{e^{3x}}{8}$
D) $y = \left( c_{1} + c_{2}x \right)e^{x} + \frac{e^{- 3x}}{8}$
Explanation
(c)
The given equation can be written as
$\left( D^{2} + 2D + 1 \right)y = 2e^{3x},\ \ where\ \ \frac{d}{dx} = D$
Here, $F(D) = D^{2} + 2D + 1\ \ \ \ \ \ and\ \ \ \ Q = 2e^{3x}$
The auxiliary equation is
$m^{2} + 2m + 1 = 0$
$\Rightarrow$ $(m + 1)^{2} = 0$
$\Rightarrow$ $m = - 1, - 1$
$\therefore$ The $CF = \left( c_{1} + c_{2}x \right)e^{- 1}$
and $PI = \frac{1}{F(D)}2e^{3x} = 2\frac{1}{D^{2} + 2D + 1}\ e^{3x}$
$= 2\frac{e^{3x}}{9 + 6 + 1} = \frac{e^{3x}}{8}$
$\therefore$ The complete solution is
$y = \left( c_{1} + c_{2}x \right)e^{- x} + \frac{e^{3x}}{8}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.