What is the solution for y^{''}+4y^'+4y=12t+16 ?

The solution for y^{''}+4y^'+4y=12t+16 is y=c_{1}e^{-2t}+c_{2}te^{-2t}+3t+1

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y^{''}+4y^'+4y=12t+16

y^{^{''}} + 4 y^{^{'}} + 4 y = 12 t + 16

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 4 y = 12 t + 16

Solve linear ODE:

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

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

\int_{0}^{10} 1 + 16 x^{2} d x

\int (3 + x^{2})^{\frac{3}{2}} d x

\int_{- π}^{π} cos (2 x) cos (3 x) d x

x \to 5 lim (\frac{2}{5 - x})

\frac{d}{d x} ((1 + \frac{8}{x})^{x})

What is the solution for y^{''}+4y^'+4y=12t+16 ?
The solution for y^{''}+4y^'+4y=12t+16 is y=c_{1}e^{-2t}+c_{2}te^{-2t}+3t+1