What is the solution for y^{''}+4y^'+4y=-3e^{2t} ?

The solution for y^{''}+4y^'+4y=-3e^{2t} is y=c_{1}e^{-2t}+c_{2}te^{-2t}-3/16 e^{2t}

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y^{''}+4y^'+4y=-3e^{2t}

y^{^{''}} + 4 y^{^{'}} + 4 y = - 3 e^{2 t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} - \frac{3}{16} e^{2 t}

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = - 3 e^{2 t}

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} - \frac{3}{16} e^{2 t}

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} - \frac{3}{16} e^{2 t}

n = 1 \sum \infty \frac{1}{2 ^{n + 2}}

f (x) = 18 e^{x} + 10 x, at x = 0

\int \frac{6 x}{x ^{3} - 8} d x

\int sec^{2} (π x) d x

\int_{- 1}^{1} t (1 - t)^{2} d t

What is the solution for y^{''}+4y^'+4y=-3e^{2t} ?
The solution for y^{''}+4y^'+4y=-3e^{2t} is y=c_{1}e^{-2t}+c_{2}te^{-2t}-3/16 e^{2t}