What is the solution for y^{''}+14y^'+50y=0 ?

The solution for y^{''}+14y^'+50y=0 is y=e^{-7t}(c_{1}cos(t)+c_{2}sin(t))

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y^{''}+14y^'+50y=0

y^{^{''}} + 14 y^{^{'}} + 50 y = 0

y = e^{- 7 t} (c_{1} cos (t) + c_{2} sin (t))

Solution steps

y^{''} (t) + 14 y^{'} (t) + 50 y = 0

Solve linear ODE:

y = e^{- 7 t} (c_{1} cos (t) + c_{2} sin (t))

y = e^{- 7 t} (c_{1} cos (t) + c_{2} sin (t))

\int (\frac{x}{( x ^{2} + 1 ) ^{2}}) d x

x y^{^{'}} + y = 3 x y^{2}, y (1) = - 3

7 x

\frac{\partial}{\partial y} (- 4 x^{3} y)

\frac{d}{d x} (3 d x)

What is the solution for y^{''}+14y^'+50y=0 ?
The solution for y^{''}+14y^'+50y=0 is y=e^{-7t}(c_{1}cos(t)+c_{2}sin(t))