What is the solution for y^{''}+2y^'+6y=0 ?

The solution for y^{''}+2y^'+6y=0 is y=e^{-t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))

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y^{''}+2y^'+6y=0

y^{^{''}} + 2 y^{^{'}} + 6 y = 0

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 6 y = 0

Solve linear ODE:

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

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

y^{^{''}} + 25 y = 20 sin (5 t)

2 s

(2 x + y) d x - x d y = 0

\frac{\partial}{\partial y} (x + y)

(x^{2} y^{4} + x^{6}) d x - x^{3} y^{3} d y = 0

What is the solution for y^{''}+2y^'+6y=0 ?
The solution for y^{''}+2y^'+6y=0 is y=e^{-t}(c_{1}cos(sqrt(5)t)+c_{2}sin(sqrt(5)t))