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

The solution for 2y^{''}+y^'-y=x+5 is y=c_{1}e^{x/2}+c_{2}e^{-x}-x-6

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

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

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- x} - x - 6

Solution steps

2 y^{''} (x) + y^{'} (x) - y = x + 5

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- x} - x - 6

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- x} - x - 6

(x + y)^{2} d x + (2 x y + x^{2} - 8) d y = 0, y (1) = 1

\frac{d}{d x} (ln (\frac{x - 2}{x + 2}))

3 y^{^{''}} - 10 y^{^{'}} + 3 y = 0, y (0) = - 5, y^{^{'}} (0) = - 1

\int_{0}^{π} sin^{2} (x) cos (x) d x

f (x) = (6 x - 5)^{\frac{1}{2}}, at x = 1

What is the solution for 2y^{''}+y^'-y=x+5 ?
The solution for 2y^{''}+y^'-y=x+5 is y=c_{1}e^{x/2}+c_{2}e^{-x}-x-6