What is the solution for y^{''}+2y^'-3y=1+xe^x ?

The solution for y^{''}+2y^'-3y=1+xe^x is y=c_{1}e^x+c_{2}e^{-3x}-1/3+(e^xx^2)/8-(e^xx)/(16)

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y^{''}+2y^'-3y=1+xe^x

y^{^{''}} + 2 y^{^{'}} - 3 y = 1 + x e^{x}

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{1}{3} + \frac{e ^{x} x ^{2}}{8} - \frac{e ^{x} x}{16}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{1}{3} + \frac{e ^{x} x ^{2}}{8} - \frac{e ^{x} x}{16}

y = c_{1} e^{x} + c_{2} e^{- 3 x} - \frac{1}{3} + \frac{e ^{x} x ^{2}}{8} - \frac{e ^{x} x}{16}

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

4 y^{^{''}} - y = e^{\frac{x}{2}} + 5

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

y^{^{'''}} - 6 y^{^{''}} = 2 - cos (x)

y^{^{''}} + 7 y^{^{'}} - 12 y = e^{3 x} - 2 e^{x} - 5 x

What is the solution for y^{''}+2y^'-3y=1+xe^x ?
The solution for y^{''}+2y^'-3y=1+xe^x is y=c_{1}e^x+c_{2}e^{-3x}-1/3+(e^xx^2)/8-(e^xx)/(16)