What is the solution for y^{''}+2y^'+36y=0 ?

The solution for y^{''}+2y^'+36y=0 is y=e^{-t}(c_{1}cos(sqrt(35)t)+c_{2}sin(sqrt(35)t))

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y^{''}+2y^'+36y=0

y^{^{''}} + 2 y^{^{'}} + 36 y = 0

y = e^{- t} (c_{1} cos (35 t) + c_{2} sin (35 t))

Solution steps

y^{''} (t) + 2 y^{'} (t) + 36 y = 0

Solve linear ODE:

y = e^{- t} (c_{1} cos (35 t) + c_{2} sin (35 t))

y = e^{- t} (c_{1} cos (35 t) + c_{2} sin (35 t))

x^{3} y^{^{''}} + 2 x^{3} y^{^{'}} + x^{3} y = 0

16 y^{^{''}} - 6 y^{^{'}} - 7 y = 0

y^{^{'''}} - y^{^{''}} + 3 y^{^{'}} + 5 y = 0

y^{^{''}} + 4 y^{^{'}} + 2 y = 0, y (0) = 0, y^{^{'}} (0) = 2

y^{^{''}} + 20 y^{^{'}} + 36 y = 0, y (0) = 1, y^{^{'}} (0) = 0

What is the solution for y^{''}+2y^'+36y=0 ?
The solution for y^{''}+2y^'+36y=0 is y=e^{-t}(c_{1}cos(sqrt(35)t)+c_{2}sin(sqrt(35)t))