Question

A function $y\left(t\right)$, such that $y\left(0\right)=1$ and $y\left(1\right)=3e^{-1}$, is a solution of the differential equation $\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}+y=0$. Then $y\left(2\right)$ is

• A. $5e^{-1}$
• B. $5e^{-2}$
• C. $7e^{-1}$
• D. $7e^{-2}$

Answer 1

The correct option is B $5e^{-2}$

$\frac{d^{2}y}{d{t}^2}+2\frac{dy}{dt}+y=0$ .... $\left(i\right)$
Auxiliary equation (A.E)
$D^2+2D+1=0$
$\Rightarrow$ $D=-1,-1\to$ two same roots
$\therefore$ The complete solution of equation (i) is
$y=(C_1+C_{2}x)e^{-x}$
Given, $y\left(0\right)=1$ and $y\left(1\right)=3e^{-1}$
$\Rightarrow$ $1=C_1,3e^{-1}=(1+C_2)e^{-1}$
$\therefore$ $y\left(x\right)=(1+2x)e^{-x}$
$\Rightarrow$ $y\left(2\right)=(1+2\times 2)e^{-2}=5e^{-2}$