What is the solution for y^{''}+4y^'+3y=te^{-t} ?

The solution for y^{''}+4y^'+3y=te^{-t} is y=c_{1}e^{-t}+c_{2}e^{-3t}+(e^{-t}t^2)/4-(e^{-t}t)/4

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y^{''}+4y^'+3y=te^{-t}

y^{^{''}} + 4 y^{^{'}} + 3 y = t e^{- t}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{e ^{- t} t ^{2}}{4} - \frac{e ^{- t} t}{4}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 3 y = t e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{e ^{- t} t ^{2}}{4} - \frac{e ^{- t} t}{4}

y = c_{1} e^{- t} + c_{2} e^{- 3 t} + \frac{e ^{- t} t ^{2}}{4} - \frac{e ^{- t} t}{4}

x \to 6 lim (\frac{3 x}{x - 6})

\int r^{2} e^{r} d r

\frac{d}{d x} (cos (arctan (x)))

\int 6 e^{2 x} d x

4 cos (t) - 9 sin (t)

What is the solution for y^{''}+4y^'+3y=te^{-t} ?
The solution for y^{''}+4y^'+3y=te^{-t} is y=c_{1}e^{-t}+c_{2}e^{-3t}+(e^{-t}t^2)/4-(e^{-t}t)/4