What is the solution for y^{''}+5y^'+6y=20sin(2t) ?

The solution for y^{''}+5y^'+6y=20sin(2t) is y=c_{1}e^{-2t}+c_{2}e^{-3t}+5/13 sin(2t)-25/13 cos(2t)

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y^{''}+5y^'+6y=20sin(2t)

y^{^{''}} + 5 y^{^{'}} + 6 y = 20 sin (2 t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + \frac{5}{13} sin (2 t) - \frac{25}{13} cos (2 t)

Solution steps

y^{''} (t) + 5 y^{'} (t) + 6 y = 20 sin (2 t)

Solve linear ODE:

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + \frac{5}{13} sin (2 t) - \frac{25}{13} cos (2 t)

y = c_{1} e^{- 2 t} + c_{2} e^{- 3 t} + \frac{5}{13} sin (2 t) - \frac{25}{13} cos (2 t)

y^{^{''}} - 2 y^{^{'}} + y = 25 sin (2 x)

y^{^{''}} - 8 y^{^{'}} + 20 y = 200 x^{2} - 65 x e^{x}

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

y^{^{'''}} - y^{^{''}} - 8 y^{^{'}} + 12 y = 2 e^{t}

y^{^{''}} + 4 y = 3 x^{3}

What is the solution for y^{''}+5y^'+6y=20sin(2t) ?
The solution for y^{''}+5y^'+6y=20sin(2t) is y=c_{1}e^{-2t}+c_{2}e^{-3t}+5/13 sin(2t)-25/13 cos(2t)