What is the solution for y^{''}+4y^'+5y=t ?

The solution for y^{''}+4y^'+5y=t is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+t/5-4/25

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y^{''}+4y^'+5y=t

y^{^{''}} + 4 y^{^{'}} + 5 y = t

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{t}{5} - \frac{4}{25}

Solution steps

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

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{t}{5} - \frac{4}{25}

y = e^{- 2 t} (c_{1} cos (t) + c_{2} sin (t)) + \frac{t}{5} - \frac{4}{25}

y^{^{''}} - y^{^{'}} - 2 y = 4 x e^{3 x}

y^{^{''}} + 9 y = cos (t) \cdot H (t - π)

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

y^{''} - 25 y = x^{2} + x

D, D (x) = \frac{45}{x + 3}

What is the solution for y^{''}+4y^'+5y=t ?
The solution for y^{''}+4y^'+5y=t is y=e^{-2t}(c_{1}cos(t)+c_{2}sin(t))+t/5-4/25