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}}

10 x^{^{''}} + 30 x^{^{'}} + 20 x = 0, x (0) = - 7, x^{^{'}} (0) = - 2

y^{^{''}} - 14 y^{^{'}} - 45 y = 0

y^{^{''}} - (2 a - 3) y^{^{'}} + a (a - 3) y = 0

5 y^{^{''}} + 20 y^{^{'}} + 640 y = 0

y^{^{''}} - 19 y^{^{'}} + 10 y = 0

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}