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

Question
Differential equations
Solve the Differential equations
$$\displaystyle{\frac{{{\left({d}^{{{2}}}\right)}{y}}}{\rbrace}}{\left\lbrace{d}{\left({t}^{{{2}}}\right\rbrace}+{4}{\left({\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}+{3}{y}={e}^{{-{t}}}\right.}\right.}$$

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
PSKypi=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-(D/2)+...)*^-1*1 =e^-t(1/2D)*1 =t(e^-t)/2ZSK
Therefore the solition is
$$\displaystyle{y}={y}{c}{f}+{y}\pi={c}{1}{\left({e}^{{-{3}{x}}}\right)}+{c}{2}{\left({e}^{{-{x}}}\right)}+\frac{{{t}{e}^{{-{t}}}}}{{2}}$$

### Relevant Questions

Solve the equation:
$$\displaystyle{\left({x}+{1}\right)}{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={x}{\left({y}^{{2}}+{1}\right)}$$
Solve the equation:
$$\displaystyle{\left({a}-{x}\right)}{\left.{d}{y}\right.}+{\left({a}+{y}\right)}{\left.{d}{x}\right.}={0}$$
Solve the following differential equations:
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}-{e}^{{-{x}}}}}{{{y}+{e}^{{{y}}}}}}$$
Solve the given system of differential equations.
\[Dx+Dy+(D+1)z=0\)
Dx+y=e^{t}\)
Dx+y-2z=50\sin(2t)\)
Solve
$$\left(d^{2}\frac{y}{dt^{2}}\right)\ +\ 7\left(\frac{dy}{dt}\right)\ +\ 10y=4te^{-3}t$$ with
$$y(0)=0,\ y'(0)=\ -1$$
Solve the following differential equation by using linear equations.
$$dx/dt = 1- t + x - tx$$
Solve the differential equations
(1) $$\displaystyle{x}{y}'-{2}{y}={x}^{{3}}{e}^{{x}}$$
(2) $$\displaystyle{\left({2}{y}{\left.{d}{x}\right.}+{\left.{d}{y}\right.}\right)}{e}^{{2}}{x}={0}$$
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{t}\right.}}}}={\frac{{{3}{t}^{{{2}}}}}{{{y}}}}$$
Solve $$d\frac{d^{2}y}{dt^{2}}\ -\ 8d\frac{dy}{dt}\ +\ 15y=9te^{3t}\ with\ y(0)=5,\ y'(0)=10$$