What is the solution for 2y^{''}+2y^'+7y=0 ?

The solution for 2y^{''}+2y^'+7y=0 is y=e^{-t/2}(c_{1}cos((sqrt(13)t)/2)+c_{2}sin((sqrt(13)t)/2))

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2y^{''}+2y^'+7y=0

2 y^{^{''}} + 2 y^{^{'}} + 7 y = 0

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2}))

Solution steps

2 y^{''} (t) + 2 y^{'} (t) + 7 y = 0

Solve linear ODE:

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2}))

y = e^{- \frac{t}{2}} (c_{1} cos (\frac{13 t}{2}) + c_{2} sin (\frac{13 t}{2}))

y^{^{''''}} + k^{2} y^{^{''}} = 0

\frac{d ^{2} y}{d x ^{2}} - k^{2} y = 0

\frac{d ^{2} y}{d x ^{2}} + 4 \frac{d y}{d x} + 3 y = 0

y^{^{'''}} - 4 y^{^{''}} + 5 y^{^{'}} - 2 y = 0

2 y^{^{''''}} + 11 y^{^{'''}} + 18 y^{^{''}} + 4 y^{^{'}} - 8 y = 0

What is the solution for 2y^{''}+2y^'+7y=0 ?
The solution for 2y^{''}+2y^'+7y=0 is y=e^{-t/2}(c_{1}cos((sqrt(13)t)/2)+c_{2}sin((sqrt(13)t)/2))