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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} + y^{^{'}} - 2 y = 4 e^{2 t}, y (0) = 1, y^{^{'}} (0) = - 1

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

y^{^{''}} - 14 y^{^{'}} + 49 y = 21 x + 6

y^{^{''}} + 2 y^{^{'}} + 2 y = x + cos (x)

D (t) = 30 e^{- 0.1 (3)}

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