What is the solution for y^{''}+4y^'+5y=1 ?

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

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y^{''}+4y^'+5y=1

y^{^{''}} + 4 y^{^{'}} + 5 y = 1

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 5 y = 1

Solve linear ODE:

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

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

(D^{3} - 3 D^{2} + 4 D - 2) y = cos (x)

\frac{d ^{2} y}{d t ^{2}} - 3 \frac{d y}{d t} + 4 y = cos (2 t)

y^{^{''}} + 5 y^{^{'}} = 100 sin (5 t) + 150 cos (5 t)

y^{^{''}} + 49 y = - 5 cos (7 t + π^{3})

y^{^{''}} + b^{2} y = sin (b t), y (0) = 0, y^{^{'}} (0) = 0

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