What is the solution for 2y^{''}+y^'-y=x+7,y(0)=1,y^'(0)=0 ?

The solution for 2y^{''}+y^'-y=x+7,y(0)=1,y^'(0)=0 is y=(-7/3+9)e^{x/2}+7/3 e^{-x}-x-8

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2y^{''}+y^'-y=x+7,y(0)=1,y^'(0)=0

2 y^{^{''}} + y^{^{'}} - y = x + 7, y (0) = 1, y^{^{'}} (0) = 0

y = (- \frac{7}{3} + 9) e^{\frac{x}{2}} + \frac{7}{3} e^{- x} - x - 8

Solution steps

2 y^{''} (x) + y^{'} (x) - y = x + 7

Solve linear ODE:

y = (- \frac{7}{3} + 9) e^{\frac{x}{2}} + \frac{7}{3} e^{- x} - x - 8

y = (- \frac{7}{3} + 9) e^{\frac{x}{2}} + \frac{7}{3} e^{- x} - x - 8

y^{^{''}} - 4 y^{^{'}} = 6 e^{3 t} - 3 e^{- t}

y^{^{''}} + 2 y^{^{'}} - 24 y = 16 - (x + 2) e^{4 x}

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

2 y^{^{''}} + 18 y = 6 tan (3 x)

y^{^{''}} + a y^{^{'}} = b

What is the solution for 2y^{''}+y^'-y=x+7,y(0)=1,y^'(0)=0 ?
The solution for 2y^{''}+y^'-y=x+7,y(0)=1,y^'(0)=0 is y=(-7/3+9)e^{x/2}+7/3 e^{-x}-x-8