What is the solution for 1/4 q^{''}+20q^'+300q=0 ?

The solution for 1/4 q^{''}+20q^'+300q=0 is q(t)=c_{1}e^{-20t}+c_{2}e^{-60t}

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1/4 q^{''}+20q^'+300q=0

\frac{1}{4} q^{^{''}} + 20 q^{^{'}} + 300 q = 0

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

Solution steps

\frac{1}{4} q^{''} (t) + 20 q^{'} (t) + 300 q (t) = 0

Solve linear ODE:

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

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

\int 2^{X} d X

\frac{2}{x + 1}

\frac{\partial}{\partial x} ((x + 2 y^{3})^{\frac{1}{3}})

\frac{d}{d x} (\frac{( x cos ( x ) + π )}{x - π})

x^{- 0.5}

What is the solution for 1/4 q^{''}+20q^'+300q=0 ?
The solution for 1/4 q^{''}+20q^'+300q=0 is q(t)=c_{1}e^{-20t}+c_{2}e^{-60t}