What is the solution for 0.5q^{''}+20q^'+1/(0.0025)q=100 ?

The solution for 0.5q^{''}+20q^'+1/(0.0025)q=100 is q(t)=e^{-20t}(c_{1}cos(20t)+c_{2}sin(20t))+0.25

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0.5q^{''}+20q^'+1/(0.0025)q=100

0.5 q^{^{''}} + 20 q^{^{'}} + \frac{1}{0.0025} q = 100

q (t) = e^{- 20 t} (c_{1} cos (20 t) + c_{2} sin (20 t)) + 0.25

Solution steps

0.5 q^{''} (t) + 20 q^{'} (t) + \frac{1}{0.0025} q (t) = 100

Solve linear ODE:

q (t) = e^{- 20 t} (c_{1} cos (20 t) + c_{2} sin (20 t)) + 0.25

q (t) = e^{- 20 t} (c_{1} cos (20 t) + c_{2} sin (20 t)) + 0.25

What is the solution for 0.5q^{''}+20q^'+1/(0.0025)q=100 ?
The solution for 0.5q^{''}+20q^'+1/(0.0025)q=100 is q(t)=e^{-20t}(c_{1}cos(20t)+c_{2}sin(20t))+0.25