What is the solution for 2y^{''}+y^'-y=3x+4 ?

The solution for 2y^{''}+y^'-y=3x+4 is y=c_{1}e^{x/2}+c_{2}e^{-x}-3x-7

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2y^{''}+y^'-y=3x+4

2 y^{^{''}} + y^{^{'}} - y = 3 x + 4

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- x} - 3 x - 7

Solution steps

2 y^{''} (x) + y^{'} (x) - y = 3 x + 4

Solve linear ODE:

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- x} - 3 x - 7

y = c_{1} e^{\frac{x}{2}} + c_{2} e^{- x} - 3 x - 7

y^{^{''}} - 5 y^{^{'}} + 6 y = x^{2} - 4

y^{^{''}} + 2 y^{^{'}} + y = x e^{- x} + 4

y^{^{''}} + 2 y = - 8 x sin (3 x)

y^{^{''}} - 14 y^{^{'}} + 49 y = 10 e^{7 x}

y^{^{''}} + y^{^{'}} - 12 y = sin (t), y (0) = 1, y^{^{'}} (0) = 1

What is the solution for 2y^{''}+y^'-y=3x+4 ?
The solution for 2y^{''}+y^'-y=3x+4 is y=c_{1}e^{x/2}+c_{2}e^{-x}-3x-7