What is the solution for y^{''}+20y^'+200y=320 ?

The solution for y^{''}+20y^'+200y=320 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+8/5

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y^{''}+20y^'+200y=320

y^{^{''}} + 20 y^{^{'}} + 200 y = 320

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{8}{5}

Solution steps

y^{''} (t) + 20 y^{'} (t) + 200 y = 320

Solve linear ODE:

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{8}{5}

y = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{8}{5}

y^{^{'''}} - 2 y^{^{'}} + 4 y = sin (2 x)

y^{^{''}} + 9 y^{^{'}} + 14 y = 3 sin (2 x)

x^{^{''}} - x = e^{- t}

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

y^{^{''}} + 4 y^{^{'}} + 6 y = 2 e^{- 3 x}

What is the solution for y^{''}+20y^'+200y=320 ?
The solution for y^{''}+20y^'+200y=320 is y=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+8/5