Question

The solution of the differential equation $\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}+y=0$ with $y\left(0\right)=y^{\prime }\left(0\right)=1$ is

• A. $(2-t)e^t$
• B. $(1+2t)e^{-t}$
• C. $(2+t)e^{-t}$
• D. $(1-2t)e^t$

Answer 1

The correct option is B $(1+2t)e^{-t}$

$\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}+y=0,y\left(0\right)=y^{\prime }\left(0\right)=1$
A.E. is $D^2+2D+1=0$
$D=-1,-1$
$y=(C_1+C_{2}t)e^{-t}$
$y\left(0\right)=1$
$\Rightarrow$ $1=C_1$
$y^{\prime }\left(t\right)=-(C_1+C_{2}t)e^{-t}+C_2{e}^{-t}$
$y^{\prime }\left(0\right)=1$
$\Rightarrow$ $1=-C_1+C_2$
$C_2=1+C_1=2$
$y=(1+2t)e^{-t}$