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

The solution for 2y^{''}+2y^'-2y=0 is y=c_{1}e^{((-1+sqrt(5))t)/2}+c_{2}e^{-((1+sqrt(5))t)/2}

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

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

y = c_{1} e^{\frac{( - 1 + 5 ) t}{2}} + c_{2} e^{- \frac{( 1 + 5 ) t}{2}}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{\frac{( - 1 + 5 ) t}{2}} + c_{2} e^{- \frac{( 1 + 5 ) t}{2}}

y = c_{1} e^{\frac{( - 1 + 5 ) t}{2}} + c_{2} e^{- \frac{( 1 + 5 ) t}{2}}

2 y^{^{''}} + 36 y^{^{'}} + 163 y = 0

\frac{d ^{2} y}{d x ^{2}} + 8 \frac{d y}{d x} + 25 y = 0

y^{^{'''}} - 3 y^{^{''}} + 4 y^{^{'}} - 2 y = 0

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

y^{^{''}} + 9 y = 0, y (0) = 0, y^{^{'}} (0) = 3

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