What is the solution for y^{''}+20y^'+200y=400 ?

The solution for y^{''}+20y^'+200y=400 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+2

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y^{''}+20y^'+200y=400

y^{^{''}} + 20 y^{^{'}} + 200 y = 400

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

Solution steps

y^{''} (t) + 20 y^{'} (t) + 200 y = 400

Solve linear ODE:

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

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

y^{^{'''}} - 7 y^{^{''}} = x^{2} + 2 cos (x)

y^{^{''}} + y = tan^{2} (t)

y^{^{''''}} + 2 y^{^{''}} + y = 9 + cos (6 t)

y^{^{''}} + y^{^{'}} - 2 y = 2 + 5 cos (2 x)

y^{^{'''}} + 4 y^{^{'}} = tan (x)

What is the solution for y^{''}+20y^'+200y=400 ?
The solution for y^{''}+20y^'+200y=400 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+2