What is the solution for y^{''}+2y^'-5y=-2 ?

The solution for y^{''}+2y^'-5y=-2 is y=c_{1}e^{(-1+sqrt(6))t}+c_{2}e^{(-1-sqrt(6))t}+2/5

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y^{''}+2y^'-5y=-2

y^{^{''}} + 2 y^{^{'}} - 5 y = - 2

y = c_{1} e^{(- 1 + 6) t} + c_{2} e^{(- 1 - 6) t} + \frac{2}{5}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{(- 1 + 6) t} + c_{2} e^{(- 1 - 6) t} + \frac{2}{5}

y = c_{1} e^{(- 1 + 6) t} + c_{2} e^{(- 1 - 6) t} + \frac{2}{5}

4 y^{^{''}} + 16 y = 1, y (0) = - 2, y^{^{'}} (0) = 0

y^{^{''}} + y = t^{3}

y^{^{''}} - 8 y^{^{'}} + 16 y = x

y^{^{''}} - 6 y^{^{'}} + 9 y = e^{x}

y^{^{''}} - 10 y^{^{'}} + 25 y = 6 t^{2} e^{5 t}

What is the solution for y^{''}+2y^'-5y=-2 ?
The solution for y^{''}+2y^'-5y=-2 is y=c_{1}e^{(-1+sqrt(6))t}+c_{2}e^{(-1-sqrt(6))t}+2/5