Question

The solution of the differential equation
$\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}-2y=3e^{2x}$, where, $y\left(0\right)=0$ and $y\left(0\right)=-2$

• A. $y=e^{-x}-e^{2x}+x{e}^{2x}$
• B. $y=e^x-e^{-2x}-x{e}^{2x}$
• C. $y=e^{-x}+e^{2x}+x{e}^{2x}$
• D. $y=e^x-e^{-2x}+x{e}^{2x}$

Answer 1

The correct option is A $y=e^{-x}-e^{2x}+x{e}^{2x}$

Given DE is
$\frac{d^{2}y}{d{x}^2}-\frac{dy}{dx}-2y=3e^{2x}$
$(D^2-D-2)y=3e^{2x}$ .... (i)
$AE$ in $m^2-m-2=0$
$\Rightarrow (m+1)(m-2)=0$
$\Rightarrow m=-1and2$
So, $CF=C_1{e}^{-x}+C_2{e}^{2x}$
Now $PI=\frac{1}{f\left(D\right)}Q=\frac{1}{D^2-D-2}\left(3e^{2x}\right)$
$3\left[\frac{x}{2D-1}\right(e^{2x}\left)\right]=3\left[\frac{x}{2\times 2-1}\right(e^{2x}\left)\right]=x{e}^{2x}$
General solution is $y=CF+PI$
$y=C_1{e}^{-x}+C_2{e}^{2x}+x{e}^{2x}$
$\frac{dy}{dx}=y^{\prime }=-C_1{e}^{-x}+2C_2{e}^{2x}+2x{e}^{2x}+e^{2x}$
using $y\left(0\right)=0\Rightarrow 0=C_1+C_2$
using $y^{\prime }\left(0\right)=-2\Rightarrow -2=-C_1+2C_2+0+1$
or $-C_1+2C_2=-3$
on solving $C_2=-1,C_1=1$
Solution is
$y=e^{-x}-e^{2x}+x{e}^{2x}$