What is the solution for y^{''}+10y^'+27y=0 ?

The solution for y^{''}+10y^'+27y=0 is y=e^{-5t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))

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y^{''}+10y^'+27y=0

y^{^{''}} + 10 y^{^{'}} + 27 y = 0

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

Solution steps

y^{''} (t) + 10 y^{'} (t) + 27 y = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial v} (v^{2})

f (x) = \frac{25}{x}

\int 5 sin (x) d x

\frac{d}{d x} (x^{2} + y)

\frac{d}{d y} (\frac{t}{2 y ^{4}})

What is the solution for y^{''}+10y^'+27y=0 ?
The solution for y^{''}+10y^'+27y=0 is y=e^{-5t}(c_{1}cos(sqrt(2)t)+c_{2}sin(sqrt(2)t))