What is the solution for 2y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=-5 ?

The solution for 2y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=-5 is y=e^{-t/2}(cos((3t)/2)-3sin((3t)/2))

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2y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=-5

2 y^{^{''}} + 2 y^{^{'}} + 5 y = 0, y (0) = 1, y^{^{'}} (0) = - 5

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

Solution steps

2 y^{''} (t) + 2 y^{'} (t) + 5 y = 0

Solve linear ODE:

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

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

y^{^{'''}} + 3 y^{^{''}} - 54 y = 0

y^{^{'''}} - 3 y^{^{''}} = 0

y^{^{'''}} - 6 y^{^{''}} - 36 y^{^{'}} + 216 y = 0

4 y^{^{''}} - 8 y^{^{'}} + 5 y = 0

y^{^{''}} - 2 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = 5

What is the solution for 2y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=-5 ?
The solution for 2y^{''}+2y^'+5y=0,y(0)=1,y^'(0)=-5 is y=e^{-t/2}(cos((3t)/2)-3sin((3t)/2))