What is the solution for y^{''}+2y^'+y=-4e^{-t} ?

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

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y^{''}+2y^'+y=-4e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = - 4 e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} - 2 t^{2} e^{- t}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = - 4 e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} - 2 t^{2} e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} - 2 t^{2} e^{- t}

y^{^{''}} + 9 y = 4 x e^{(2 x)}, y (0) = 0, y^{^{'}} (0) = 0

x \to 0 + lim (5 x + 9)

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

\frac{\partial}{\partial x} (\frac{x ^{2}}{y ^{2}})

\int_{- 1}^{2} \frac{x}{x + 2} d x

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