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

The solution for 2y^{''}+y^'-3y=e^x-x is y=c_{1}e^x+c_{2}e^{-(3x)/2}+(e^xx)/5+x/3+1/9

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

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

y = c_{1} e^{x} + c_{2} e^{- \frac{3 x}{2}} + \frac{e ^{x} x}{5} + \frac{x}{3} + \frac{1}{9}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- \frac{3 x}{2}} + \frac{e ^{x} x}{5} + \frac{x}{3} + \frac{1}{9}

y = c_{1} e^{x} + c_{2} e^{- \frac{3 x}{2}} + \frac{e ^{x} x}{5} + \frac{x}{3} + \frac{1}{9}

y^{^{''}} + 4 y = \frac{1}{sin ( 2 t )}

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

y^{^{''}} - 4 y^{^{'}} + 4 y = - 7 x + 4 x^{2}

0.5 y^{^{''}} + y^{^{'}} + 4 y = 10 sin (2 x)

y^{^{''}} - y^{^{'}} = \frac{e ^{x}}{e ^{x} + 1}

What is the solution for 2y^{''}+y^'-3y=e^x-x ?
The solution for 2y^{''}+y^'-3y=e^x-x is y=c_{1}e^x+c_{2}e^{-(3x)/2}+(e^xx)/5+x/3+1/9