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

The solution for 2y^{''}+2y^'-y=0 is y=c_{1}e^{((-1+sqrt(3))t)/2}+c_{2}e^{-((1+sqrt(3))t)/2}

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

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

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

Solution steps

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

Solve linear ODE:

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

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

3 y^{^{''}} + 27 y = 0, y (0) = 0, y^{^{'}} (0) = 1

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

9 y^{^{''}} - 6 y^{^{'}} = 0

y^{^{'''}} + 4 y^{^{''}} = 0

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 2

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