What is the solution for y^{''}=-y^'-y ?

The solution for y^{''}=-y^'-y is y=e^{-t/2}(c_{1}cos((sqrt(3)t)/2)+c_{2}sin((sqrt(3)t)/2))

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y^{''}=-y^'-y

y^{^{''}} = - y^{^{'}} - y

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

Solution steps

y^{''} (t) = - y^{'} (t) - y

Solve linear ODE:

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

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

y^{^{''}} - 3 y^{^{'}} - 4 y = 0, y (0) = 0, y^{^{'}} (0) = 1

y^{^{''}} + y^{^{'}} = 0, y (\frac{π}{4}) = 0, y^{^{'}} (\frac{π}{4}) = - 3

y^{^{''}} + (\frac{v}{a}) y^{^{'}} = 0

\frac{d ^{4} y}{d x ^{4}} - 2 (\frac{d ^{2} y}{d x ^{2}}) + y = 0

y^{^{''}} - 16 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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