What is the solution for y^{''}+14y^'+50y=0,y(0)=2,y^'(0)=-17 ?

The solution for y^{''}+14y^'+50y=0,y(0)=2,y^'(0)=-17 is y=e^{-7t}(2cos(t)-3sin(t))

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y^{''}+14y^'+50y=0,y(0)=2,y^'(0)=-17

y^{^{''}} + 14 y^{^{'}} + 50 y = 0, y (0) = 2, y^{^{'}} (0) = - 17

y = e^{- 7 t} (2 cos (t) - 3 sin (t))

Solution steps

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

Solve linear ODE:

y = e^{- 7 t} (2 cos (t) - 3 sin (t))

y = e^{- 7 t} (2 cos (t) - 3 sin (t))

y = (2 tan^{2} (2 x) - 1)^{\frac{1}{2}}

\frac{\partial}{\partial x} (x (1 - x - 2 y))

y^{'} = 0.002 y (2500 - y)

\frac{d ^{2}}{d x ^{2}} (x^{3} + y^{3})

π^{2}

What is the solution for y^{''}+14y^'+50y=0,y(0)=2,y^'(0)=-17 ?
The solution for y^{''}+14y^'+50y=0,y(0)=2,y^'(0)=-17 is y=e^{-7t}(2cos(t)-3sin(t))