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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

9 y^{^{''}} - 4 y = sin (x)

y^{^{''}} - 3 y^{^{'}} - 4 y = 2 sin (x)

\frac{d ^{2} y}{d x ^{2}} + y = cos (x)

y^{^{''''}} + 2 y^{^{''}} + y = 3 + cos (2 t)

y^{^{''}} + 4 y = sin^{3} (x)

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