Question

The general solution of the differential equation $\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+y=2e^{3x}$ 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. $y=\left(c_1+c_{2}x\right)e^x+\frac{e^{-3x}}{8}$

Answer 1

Answer: (C)

Given : $\frac{d^{2}y}{d{x}^2}+2\frac{dy}{dx}+y=2e^{3x}$
The auxiliary equation is
$D^2+2D+1=0or{m}^2+2m+1=0$
$\Rightarrow \left(m+1\right)\left(m+1\right)=0\Rightarrow m=-1,-1$
i.e., repeated roots
$\therefore$ Complementary function = $\left(c_1+c_{2}x\right)e^{-x}$
Now Particular Integral $\left(P.I.\right)$
$=\frac{1}{D^2+2D+1}.2e^{3x}\left[D=3\right]$
P.I$=\frac{1}{3^2+2.3+1}.2.e^{3x}=\frac{2e^{3x}}{16}=\frac{e^{3x}}{8}$
Solution y = C. F. + P. I.
$\Rightarrow y=\left(c_1+c_{2}x\right)e^{-x}+\frac{e^{3x}}{8}$