What is the solution for y^{''}+4y^'+20y=0 ?

The solution for y^{''}+4y^'+20y=0 is y=e^{-2t}(c_{1}cos(4t)+c_{2}sin(4t))

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y^{''}+4y^'+20y=0

y^{^{''}} + 4 y^{^{'}} + 20 y = 0

y = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 20 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{- 2 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

\frac{\partial}{\partial x} (0.0099999)

\int 1 + 5 x^{4} (20 x^{3}) d x

n = 1 \sum \infty n (\frac{3}{2})^{n}

\frac{d}{d x} ((\frac{x ^{2} + 1}{x ^{2} - 1})^{4})

y = cos (\frac{π t}{2} - \frac{π}{3})

What is the solution for y^{''}+4y^'+20y=0 ?
The solution for y^{''}+4y^'+20y=0 is y=e^{-2t}(c_{1}cos(4t)+c_{2}sin(4t))