What is the solution for 1/2 q^{''}+10q^'+(1/(10^{-2)})q=200 ?

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

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1/2 q^{''}+10q^'+(1/(10^{-2)})q=200

\frac{1}{2} q^{^{''}} + 10 q^{^{'}} + (\frac{1}{1 0 ^{- 2}}) q = 200

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

Solution steps

\frac{1}{2} q^{''} (t) + 10 q^{'} (t) + (\frac{1}{1 0 ^{- 2}}) q (t) = 200

Solve linear ODE:

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

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

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