What is the solution for y^{''}+2y^'+y=10e^{-t} ?

The solution for y^{''}+2y^'+y=10e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+5t^2e^{-t}

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y^{''}+2y^'+y=10e^{-t}

y^{^{''}} + 2 y^{^{'}} + y = 10 e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + 5 t^{2} e^{- t}

Solution steps

y^{''} (t) + 2 y^{'} (t) + y = 10 e^{- t}

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + 5 t^{2} e^{- t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + 5 t^{2} e^{- t}

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

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

\int \frac{1}{x ^{2} + y ^{2}} d x

\frac{9}{s ^{2} + 9 s + 9} \cdot \frac{1}{s}

f (x) = x^{4} + 9 x^{2} - x, at x = 1

What is the solution for y^{''}+2y^'+y=10e^{-t} ?
The solution for y^{''}+2y^'+y=10e^{-t} is y=c_{1}e^{-t}+c_{2}te^{-t}+5t^2e^{-t}