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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

y^{^{''}} - 16 y = 96

y^{^{''}} - 14 y^{^{'}} + 49 y = 28 x + 6

y^{^{''}} - 12 y^{^{'}} + 36 y = 2 e^{6 x}

y^{^{''}} - y^{^{'}} - 6 y = 53 cos (3 x)

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

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