What is the solution for y^{''}+4*y^'+4*y=20cos(2t) ?

The solution for y^{''}+4*y^'+4*y=20cos(2t) is y=c_{1}e^{-2t}+c_{2}te^{-2t}+5/2 sin(2t)

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y^{''}+4y^'+4y=20cos(2t)

y^{^{''}} + 4 \cdot y^{^{'}} + 4 \cdot y = 20 cos (2 t)

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{5}{2} sin (2 t)

Solution steps

y^{''} (t) + 4 \cdot y^{'} (t) + 4 \cdot y = 20 cos (2 t)

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{5}{2} sin (2 t)

y = c_{1} e^{- 2 t} + c_{2} t e^{- 2 t} + \frac{5}{2} sin (2 t)

What is the solution for y^{''}+4*y^'+4*y=20cos(2t) ?
The solution for y^{''}+4*y^'+4*y=20cos(2t) is y=c_{1}e^{-2t}+c_{2}te^{-2t}+5/2 sin(2t)