What is the solution for y^{''}+2y^'+y=3-e^{-t} ?

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

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y^{''}+2y^'+y=3-e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = 3 - e^{- t}

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 3 - e^{- t}

Solve linear ODE:

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

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

\frac{d}{d x} (sec^{2} (x) + tan^{2} (x))

\int cos^{8} (x) d x

(6.4) (2.7)

(x^{- 2} + x^{- 3}) (x^{5} - 2 x^{2})

x \to - 1 lim (x^{2} + 6)

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