What is the solution for y^{''}+4y^'+9y=0 ?

The solution for y^{''}+4y^'+9y=0 is y=e^{-2t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))

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y^{''}+4y^'+9y=0

y^{^{''}} + 4 y^{^{'}} + 9 y = 0

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

Solution steps

y^{''} (t) + 4 y^{'} (t) + 9 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{x ^{2}}{196 - x ^{2}} d x

y^{^{''}} + 2 y^{^{'}} + 37 y = 0, y (0) = - 3, y^{^{'}} (0) = 21

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

\frac{d}{d x} (36 e^{- 2 x})

f (x) = - 3 e^{4 x}

What is the solution for y^{''}+4y^'+9y=0 ?
The solution for y^{''}+4y^'+9y=0 is y=e^{-2t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))