What is the solution for y^{''}+20y^'+200y=2 ?

The solution for y^{''}+20y^'+200y=2 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+1/100

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y^{''}+20y^'+200y=2

y^{^{''}} + 20 y^{^{'}} + 200 y = 2

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{1}{100}

Solution steps

y^{''} (t) + 20 y^{'} (t) + 200 y = 2

Solve linear ODE:

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{1}{100}

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{1}{100}

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

y^{^{''}} - y = 8 e^{9 x} + x e^{9 x} + 5 x^{2} e^{9 x}

y^{^{''}} + y = e^{x} + x^{3}, y (0) = 2, y^{^{'}} (0) = 0

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

y^{^{''}} - 4 y = x^{- 1} e^{2 x}

What is the solution for y^{''}+20y^'+200y=2 ?
The solution for y^{''}+20y^'+200y=2 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+1/100