Question

Consider the differential equation $\frac{d^{2}x\left(t\right)}{d{t}^2}+3\frac{dx\left(t\right)}{dt}+2x\left(t\right)=0$
Given $x\left(0\right)=20$ and $x\left(1\right)=10/e$, where $e=2.718$, the value of x(2) is

Answer 1

$\frac{d^{2}x}{d{t}^2}+3\frac{dx}{dt}+2x=0,x\left(0\right)=20,x\left(1\right)=\frac{10}{C}$
A.E. is
$D^2+3D+2=0$
$D=-1,-2$
$x=C_1{e}^{-t}+C_2{e}^{-2t}$
$x\left(0\right)=20$
$\Rightarrow$ $C_1+C_2=20$ .... $\left(1\right)$
$x\left(1\right)=\frac{10}{e}$
$\Rightarrow$ $\frac{10}{e}=\frac{C_1}{e}+\frac{C_2}{e^2}$
$\Rightarrow$ $10=C_1+\frac{C_2}{e}$ ....$\left(2\right)$
From $\left(1\right)$ and $\left(2\right)$
$C_1=\frac{10e-20}{e-1},C_2=\frac{10e}{e-1}$
$x\left(t\right)=\left(\frac{10e-20}{e-1}\right)e^{-t}+\left(\frac{10e}{e-1}\right)e^{-2t}$
$x\left(2\right)=\left(\frac{10e-20}{e-1}\right)e^{-2}+\left(\frac{10e}{e-1}\right)e^{-4}$
$=0.8556$