Q61 / 120 · VITEEE 2007 · Mathematics
The general solution of the differential equation $\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+y=2 e^{3 x}$ is given by
- A$y=\left(c_{1}+c_{2} x\right) e^{x}+\frac{e^{3 x}}{8}$
- B$y=\left(c_{1}+c_{2} x\right) e^{-x}+\frac{e^{-3 x}}{8}$
- Correct answerC$y=\left(c_{1}+c_{2} x\right) e^{-x}+\frac{e^{3 x}}{8}$
- D$y=\left(c_{1}+c_{2} x\right) e^{x}+\frac{e^{-3 x}}{8}$
Solution
Given : $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+2 \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=2 \mathrm{e}^{3 \mathrm{x}}$
The auxiliary equation is
$\mathrm{D}^{2}+2 \mathrm{D}+1=0$ or $\mathrm{m}^{2}+2 \mathrm{~m}+1=0$
$\Rightarrow(\mathrm{m}+1)(\mathrm{m}+1)=0 \Rightarrow \mathrm{m}=-1,-1$
i.e., repeated roots
$\therefore$ Complementary function $=\left(\mathrm{c}_{1}+\mathrm{c}_{2} \mathrm{x}\right) \mathrm{e}^{-\mathrm{x}}$
Now Particular Integral (P.I.)
$\begin{array}{l}
=\frac{1}{D^{2}+2 D+1} \cdot 2 e^{3 x} \quad[D=3] \\
\text { P.I. }=\frac{1}{3^{2}+2.3+1} \cdot 2 \cdot e^{3 x}=\frac{2 e^{3 x}}{16}=\frac{e^{3 x}}{8}
\end{array}$
Solution $\mathrm{y}=\mathrm{C} . \mathrm{F} .+\mathrm{P} . \mathrm{I}$
$\Rightarrow \mathrm{y}=\left(\mathrm{c}_{1}+\mathrm{c}_{2} \mathrm{x}\right) \mathrm{e}^{-\mathrm{x}}+\frac{\mathrm{e}^{3 \mathrm{x}}}{8}$
Hence, the correct answer is option: 'C'.