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^{^{'}} = (t y^{2} - 1) y, y (1) = \frac{1}{2}

\int 1 - e^{3 x} d x

f (x) = \frac{3}{x}, (3, 1)

x^{2} y^{^{'}} + 4 x y = \frac{sin ( 2 x )}{x ^{2}}

\frac{d}{d x} (C e^{- 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}