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

The solution for y^{''}+2y^'+17y=0,y^'(0)=-2 is y=e^{-t}((2+4c_{2})cos(4t)+c_{2}sin(4t))

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

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

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

Solution steps

y^{''} (t) + 2 y^{'} (t) + 17 y = 0

Solve linear ODE:

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

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

y^{^{''}} + \frac{10}{3} y^{^{'}} + a y = 0, y (0) = 1, y^{^{'}} (0) = \frac{5}{3}

u^{^{''}} + u^{^{'}} = 0

9 y^{^{''}} - 30 y^{^{'}} - 24 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + 5 y^{^{'}} + 7 y = 0

\frac{d ^{2} y}{d x ^{2}} - 7 \frac{d y}{d x} + 6 y = 0

What is the solution for y^{''}+2y^'+17y=0,y^'(0)=-2 ?
The solution for y^{''}+2y^'+17y=0,y^'(0)=-2 is y=e^{-t}((2+4c_{2})cos(4t)+c_{2}sin(4t))