What is the solution for 0.5q^{''}+10q^'+100q=150 ?

The solution for 0.5q^{''}+10q^'+100q=150 is q(t)=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+3/2

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0.5q^{''}+10q^'+100q=150

0.5 q^{^{''}} + 10 q^{^{'}} + 100 q = 150

q (t) = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{3}{2}

Solution steps

0.5 q^{''} (t) + 10 q^{'} (t) + 100 q (t) = 150

Solve linear ODE:

q (t) = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{3}{2}

q (t) = e^{- 10 t} (c_{1} cos (10 t) + c_{2} sin (10 t)) + \frac{3}{2}

\int_{0}^{\infty} 3 e^{- x} d x

12 t

30 x - 3 x

\frac{d}{d x} (\frac{x ^{2} + 4 x + 3}{x})

\frac{d}{d x} (x^{4} (tan (x) + sec (x)))

What is the solution for 0.5q^{''}+10q^'+100q=150 ?
The solution for 0.5q^{''}+10q^'+100q=150 is q(t)=e^{-10t}(c_{1}cos(10t)+c_{2}sin(10t))+3/2