Question

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

First order differential equations
ANSWERED
asked 2021-01-02

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

Answers (1)

2021-01-03

Given IVP is \(\displaystyle{y}{''}+{4}{y}'+{3}{y}={e}^{{-{t}}}.\) To find the auxiliary equation, let \(\displaystyle{y}={e}^{{{m}}}\times{x}.\) The auxiliary equation of is
\(\displaystyle{\left({m}^{{{2}}}\right)}+{4}{m}+{3}={0}\)
\(\displaystyle\Rightarrow{\left({m}+{3}\right)}{\left({m}+{1}\right)}={0}\)
\(\displaystyle\Rightarrow{m}=-{3},-{1}\)
So, the complimentary function is given by
\(\displaystyle{y}{c}{f}={c}{1}{\left({e}^{{-{3}{x}}}\right)}+{c}{2}{e}^{{-{x}}}.\)

The particular integral is given by

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

\(=e^{-t}((1/(D-1)^2)+4(D-1)+3)*1\)

\(=e^{-t}(1/(D^2)-2D+1+4D-4+3)*1\)

\(=e^{-t}((1/2D)(D/2)+1)^{-1}*1\)

\(=e^{-t}(1/2D)*1\)

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

Therefore the solition is
\(\displaystyle{y}={y}{c}{f}+{y}\pi={c}{1}{\left({e}^{{-{3}{x}}}\right)}+{c}{2}{\left({e}^{{-{x}}}\right)}+\frac{{{t}{e}^{{-{t}}}}}{{2}}\)

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