What is the solution for 2y^{''}+12y^'+50y=0 ?

The solution for 2y^{''}+12y^'+50y=0 is y=e^{-3t}(c_{1}cos(4t)+c_{2}sin(4t))

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2y^{''}+12y^'+50y=0

2 y^{^{''}} + 12 y^{^{'}} + 50 y = 0

y = e^{- 3 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

Solution steps

2 y^{''} (t) + 12 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = e^{- 3 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

y = e^{- 3 t} (c_{1} cos (4 t) + c_{2} sin (4 t))

- \frac{13}{s ^{6}}

\int e^{5 θ} d θ

\int - cos^{2} (2 x) d x

y^{^{'}} + t^{2} = 0

\int cos^{2} (2 x) sin (2 x) d x

What is the solution for 2y^{''}+12y^'+50y=0 ?
The solution for 2y^{''}+12y^'+50y=0 is y=e^{-3t}(c_{1}cos(4t)+c_{2}sin(4t))