Question

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

First order differential equations

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

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