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

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

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

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

y = (- \frac{1}{3} + 3) e^{\frac{x}{2}} + \frac{1}{3} e^{- x} - x - 2

Solution steps

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

Solve linear ODE:

y = (- \frac{1}{3} + 3) e^{\frac{x}{2}} + \frac{1}{3} e^{- x} - x - 2

y = (- \frac{1}{3} + 3) e^{\frac{x}{2}} + \frac{1}{3} e^{- x} - x - 2

\frac{d ^{2} y}{d x ^{2}} - 2 \frac{d y}{d x} = x + sin (x)

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

y^{^{''}} + y^{^{'}} = 7

y^{^{''}} - 3 y^{^{'}} - 4 y = 3 e^{2 t} + 2 sin (t)

y^{^{''}} + y^{^{'}} = 4 x

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