# Solve \frac{dx}{dt}+2x=2e^{-3t} with initial condition x(0)=2

Solve $\frac{dx}{dt}+2x=2{e}^{-3t}$ with initial condition $x\left(0\right)=2$

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Step 1
Consider the provided differential equation,
Given $\frac{dx}{dt}+2x=2{e}^{-3t}$ with initial condition $x\left(0\right)=2$
This is the ordinary differential equation.
Take Laplace on both the sides,
$L\left(\frac{dx}{dt}+2x\right)=L\left(2{e}^{-3t}\right)$
$L\left(\frac{dx}{dt}\right)+2L\left(x\right)=2L\left({e}^{-3t}\right)$

$sL\left(x\left(t\right)\right)-2+2L\left(x\left(t\right)\right)=\frac{2}{s+3}$
$L\left(x\left(t\right)\right)\left(s+2\right)=\frac{2}{s+3}+2$
$L\left(x\left(t\right)\right)\left(s+2\right)=\frac{2+2\left(s+3\right)}{s+3}$
$L\left(x\left(t\right)\right)\left(s+2\right)=\frac{2s+8}{s+3}$
Step 2
Now, further simplified the above equation,
$L\left(x\left(t\right)\right)=\frac{2s+8}{\left(s+3\right)\left(s+2\right)}$
$L\left(x\left(t\right)\right)=\frac{2\left(s+4\right)}{\left(s+3\right)\left(s+2\right)}$
$L\left(x\left(t\right)\right)=\frac{2\left[2\left(s+3\right)-\left(s-2\right)\right]}{\left(s+3\right)\left(s+2\right)}$
$L\left(x\left(t\right)\right)=\frac{4\left(s+3\right)-2\left(s+2\right)}{\left(s+3\right)\left(s+2\right)}$
$L\left(x\left(t\right)\right)=\frac{4}{\left(s+2\right)}-\frac{2}{\left(s+3\right)}$
Now ,