What is the solution for y^{''}+5y^'+4y=te^{-t} ?

The solution for y^{''}+5y^'+4y=te^{-t} is y=c_{1}e^{-t}+c_{2}e^{-4t}+(e^{-t}t^2)/6-(e^{-t}t)/9

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y^{''}+5y^'+4y=te^{-t}

y^{^{''}} + 5 y^{^{'}} + 4 y = t e^{- t}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{e ^{- t} t ^{2}}{6} - \frac{e ^{- t} t}{9}

Solution steps

y^{''} (t) + 5 y^{'} (t) + 4 y = t e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{e ^{- t} t ^{2}}{6} - \frac{e ^{- t} t}{9}

y = c_{1} e^{- t} + c_{2} e^{- 4 t} + \frac{e ^{- t} t ^{2}}{6} - \frac{e ^{- t} t}{9}

\int \frac{x ^{3} - 3 x + 2}{( x + 1 ) ( x - 1 ) ^{3}} d x

\int (\frac{2 x}{e ^{x}}) d x

\int_{0}^{\frac{π}{2}} sin (2 x - π) d x

k = 1 \sum \infty \frac{1}{k !}

y = \frac{5}{( 2 x ^{3} )} + 2 cos (x)

What is the solution for y^{''}+5y^'+4y=te^{-t} ?
The solution for y^{''}+5y^'+4y=te^{-t} is y=c_{1}e^{-t}+c_{2}e^{-4t}+(e^{-t}t^2)/6-(e^{-t}t)/9