# Solve the Differential equationsfrac{(d^{2})y)}{d(t^{2}}+4(frac{dy}{dt}+3y=e^{-t}

Solve the Differential equations
$\frac{\left({d}^{2}y\right)}{d{t}^{2}}+4\frac{dy}{dt}+3y={e}^{-t}$

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Layton

Given IVP is $y{}^{″}+4{y}^{\prime }+3y={e}^{-t}.$ To find the auxiliary equation, let $y={e}^{m}×x.$ The auxiliary equation of is
$\left({m}^{2}\right)+4m+3=0$
$⇒\left(m+3\right)\left(m+1\right)=0$
$⇒m=-3,-1$
So, the complimentary function is given by
$ycf=c1\left({e}^{-3x}\right)+c2{e}^{-x}.$

The particular integral is given by

$y\pi =1/\left(\left({D}^{2}\right)+4d+3\right){e}^{-t}$

$={e}^{-t}\left(\left(1/\left(D-1{\right)}^{2}\right)+4\left(D-1\right)+3\right)\ast 1$

$={e}^{-t}\left(1/\left({D}^{2}\right)-2D+1+4D-4+3\right)\ast 1$

$={e}^{-t}\left(\left(1/2D\right)\left(D/2\right)+1{\right)}^{-1}\ast 1$

$={e}^{-t}\left(1/2D\right)\ast 1$

$=t\left({e}^{-t}\right)/2$

Therefore the solition is
$y=ycf+y\pi =c1\left({e}^{-3x}\right)+c2\left({e}^{-x}\right)+\frac{t{e}^{-t}}{2}$