What is the solution for q^{''}+40q^'+4000q=0 ?

The solution for q^{''}+40q^'+4000q=0 is q(t)=e^{-20t}(c_{1}cos(60t)+c_{2}sin(60t))

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q^{''}+40q^'+4000q=0

q^{^{''}} + 40 q^{^{'}} + 4000 q = 0

q (t) = e^{- 20 t} (c_{1} cos (60 t) + c_{2} sin (60 t))

Solution steps

q^{''} (t) + 40 q^{'} (t) + 4000 q (t) = 0

Solve linear ODE:

q (t) = e^{- 20 t} (c_{1} cos (60 t) + c_{2} sin (60 t))

q (t) = e^{- 20 t} (c_{1} cos (60 t) + c_{2} sin (60 t))

\int \frac{17 x - 3}{3 x ^{2} + x - 2} d x

\int x^{2} x^{3} + 6 d x

y = x^{3} - 9 x, (1, - 8)

\int \frac{5 - 4 x ^{2}}{3 + 2 x} d x

y = \frac{e ^{x} + e ^{- x}}{x}

What is the solution for q^{''}+40q^'+4000q=0 ?
The solution for q^{''}+40q^'+4000q=0 is q(t)=e^{-20t}(c_{1}cos(60t)+c_{2}sin(60t))