What is the solution for y^{''}+2y^'+5y=1 ?

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

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y^{''}+2y^'+5y=1

y^{^{''}} + 2 y^{^{'}} + 5 y = 1

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 5 y = 1

Solve linear ODE:

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

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

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

x^{^{''}} + 2 x^{^{'}} + x = e^{(- t)}

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 8 y = 8 x^{2} + 8 x + 1

x^{^{''}} + 2 x^{^{'}} + 5 x = 1, x (0) = 0, x^{^{'}} (0) = 0

y^{^{''}} + 9 y = e^{- 2 x}

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