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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y = x^{2} (1 - 2 x)

\int_{0}^{4} x - x + 2 d x

x \to 1 lim (\frac{- 2}{x - 1})

\frac{d}{d x} (e^{x} + 2)

y = ln (5 x - 2)

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