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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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