What is the solution for y^{''}+20y^'+200y=1 ?

The solution for y^{''}+20y^'+200y=1 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+1/200

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y^{''}+20y^'+200y=1

y^{^{''}} + 20 y^{^{'}} + 200 y = 1

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{1}{200}

Solution steps

y^{''} (t) + 20 y^{'} (t) + 200 y = 1

Solve linear ODE:

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{1}{200}

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{1}{200}

What is the solution for y^{''}+20y^'+200y=1 ?
The solution for y^{''}+20y^'+200y=1 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+1/200