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

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

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

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

y = c_{1} e^{- t} + c_{2} t e^{- t} + 4 e^{- 2 t}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{- t} + c_{2} t e^{- t} + 4 e^{- 2 t}

y = c_{1} e^{- t} + c_{2} t e^{- t} + 4 e^{- 2 t}

x (81.9)

(1 + x)^{2} y^{^{'}} + (1 + x) y = 1

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

(t - \frac{3 π}{6}) sin (6 t)

\frac{d}{d x} (6 ln (cos (x)))

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