What is the solution for y^{''}+4y^'+13y=5 ?

The solution for y^{''}+4y^'+13y=5 is y=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))+5/13

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y^{''}+4y^'+13y=5

y^{^{''}} + 4 y^{^{'}} + 13 y = 5

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 13 y = 5

Solve linear ODE:

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

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

y^{^{''}} + 4 y^{^{'}} + y = 2

y^{^{''}} + 5 y^{^{'}} + 6 y = 20 t e^{5 t}

y^{^{''}} - 6 y^{^{'}} + 8 y = - 3 e^{2 x} - 5 x e^{- 4 x} + 3 x

y^{^{''''}} - y^{^{''}} = 3 x^{2} - 2 x

y^{^{'''}} - 2 y^{^{''}} + y^{^{'}} = 1

What is the solution for y^{''}+4y^'+13y=5 ?
The solution for y^{''}+4y^'+13y=5 is y=e^{-2t}(c_{1}cos(3t)+c_{2}sin(3t))+5/13