What is the solution for y^{''}-4y^'+13y=4+3e^t ?

The solution for y^{''}-4y^'+13y=4+3e^t is y=e^{2t}(c_{1}cos(3t)+c_{2}sin(3t))+4/13+3/10 e^t

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y^{''}-4y^'+13y=4+3e^t

y^{^{''}} - 4 y^{^{'}} + 13 y = 4 + 3 e^{t}

y = e^{2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{4}{13} + \frac{3}{10} e^{t}

Solution steps

y^{''} (t) - 4 y^{'} (t) + 13 y = 4 + 3 e^{t}

Solve linear ODE:

y = e^{2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{4}{13} + \frac{3}{10} e^{t}

y = e^{2 t} (c_{1} cos (3 t) + c_{2} sin (3 t)) + \frac{4}{13} + \frac{3}{10} e^{t}

y^{^{''}} - y = 2 (e^{x} - e^{- x})^{- 1}

y^{^{''}} + 2 y = sec (x)

y^{^{''}} + 3 y^{^{'}} - 54 y = - 174 t + 108 t^{2}

2 y^{^{''}} + 5 y^{^{'}} + 3 y = \frac{1}{4} ln (x + 1)

y^{^{''}} + 9 y = 17 sec^{2} (3 t)

What is the solution for y^{''}-4y^'+13y=4+3e^t ?
The solution for y^{''}-4y^'+13y=4+3e^t is y=e^{2t}(c_{1}cos(3t)+c_{2}sin(3t))+4/13+3/10 e^t