What is the solution for y^{''}+6y^'+15y=0 ?

The solution for y^{''}+6y^'+15y=0 is y=e^{-3t}(c_{1}cos(sqrt(6)t)+c_{2}sin(sqrt(6)t))

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y^{''}+6y^'+15y=0

y^{^{''}} + 6 y^{^{'}} + 15 y = 0

y = e^{- 3 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

Solution steps

y^{''} (t) + 6 y^{'} (t) + 15 y = 0

Solve linear ODE:

y = e^{- 3 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

y = e^{- 3 t} (c_{1} cos (6 t) + c_{2} sin (6 t))

\int (3 t^{4} - 8) t^{3} d t

\int e^{- 4 x} x^{2} d x

\int sec (x) - \frac{1}{sin ( x )} + cos (x) d x

\int_{0}^{\infty} 6 sin^{2} (x) d x

x \to - 1 lim (2 x + 6)

What is the solution for y^{''}+6y^'+15y=0 ?
The solution for y^{''}+6y^'+15y=0 is y=e^{-3t}(c_{1}cos(sqrt(6)t)+c_{2}sin(sqrt(6)t))