What is the solution for q^{''}+20q^'+500q=12 ?

The solution for q^{''}+20q^'+500q=12 is q(t)=e^{-10t}(c_{1}cos(20t)+c_{2}sin(20t))+3/125

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q^{''}+20q^'+500q=12

q^{^{''}} + 20 q^{^{'}} + 500 q = 12

q (t) = e^{- 10 t} (c_{1} cos (20 t) + c_{2} sin (20 t)) + \frac{3}{125}

Solution steps

q^{''} (t) + 20 q^{'} (t) + 500 q (t) = 12

Solve linear ODE:

q (t) = e^{- 10 t} (c_{1} cos (20 t) + c_{2} sin (20 t)) + \frac{3}{125}

q (t) = e^{- 10 t} (c_{1} cos (20 t) + c_{2} sin (20 t)) + \frac{3}{125}

x \to 1 lim (3 x^{2} - 2 x + 1)

\int cos^{2} (3 x) sin^{2} (3 x) d x

x \to 0 lim (x sin (\frac{5}{x}))

y^{^{'}} = \frac{( y + y ^{2} )}{x ^{2}}

\frac{d}{d x} (x^{\frac{3}{4}} y^{3})

What is the solution for q^{''}+20q^'+500q=12 ?
The solution for q^{''}+20q^'+500q=12 is q(t)=e^{-10t}(c_{1}cos(20t)+c_{2}sin(20t))+3/125