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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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