What is the solution for y^{''}+4y^'+29y=0 ?

The solution for y^{''}+4y^'+29y=0 is y=e^{-2t}(c_{1}cos(5t)+c_{2}sin(5t))

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y^{''}+4y^'+29y=0

y^{^{''}} + 4 y^{^{'}} + 29 y = 0

y = e^{- 2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

Solution steps

y^{''} (t) + 4 y^{'} (t) + 29 y = 0

Solve linear ODE:

y = e^{- 2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

y = e^{- 2 t} (c_{1} cos (5 t) + c_{2} sin (5 t))

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

\frac{d}{d x} (\frac{6}{( 1 - x ) ^{3}})

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

y = 10 x, y = x^{2} - 11

\int x (2 x + 1) d x

What is the solution for y^{''}+4y^'+29y=0 ?
The solution for y^{''}+4y^'+29y=0 is y=e^{-2t}(c_{1}cos(5t)+c_{2}sin(5t))