What is the solution for x^{''}+50x^'+400x=100cos(20t) ?

The solution for x^{''}+50x^'+400x=100cos(20t) is x=c_{1}e^{-10t}+c_{2}e^{-40t}+1/10 sin(20t)

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x^{''}+50x^'+400x=100cos(20t)

x^{^{''}} + 50 x^{^{'}} + 400 x = 100 cos (20 t)

x = c_{1} e^{- 10 t} + c_{2} e^{- 40 t} + \frac{1}{10} sin (20 t)

Solution steps

x^{''} (t) + 50 x^{'} (t) + 400 x = 100 cos (20 t)

Solve linear ODE:

x = c_{1} e^{- 10 t} + c_{2} e^{- 40 t} + \frac{1}{10} sin (20 t)

x = c_{1} e^{- 10 t} + c_{2} e^{- 40 t} + \frac{1}{10} sin (20 t)

What is the solution for x^{''}+50x^'+400x=100cos(20t) ?
The solution for x^{''}+50x^'+400x=100cos(20t) is x=c_{1}e^{-10t}+c_{2}e^{-40t}+1/10 sin(20t)