What is the solution for y^{''}-4y^'+3y=(x^2-3x)e^x ?

The solution for y^{''}-4y^'+3y=(x^2-3x)e^x is y=c_{1}e^{3x}+c_{2}e^x-(e^xx^3)/6+(e^xx^2)/2+(e^xx)/2

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

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

y = c_{1} e^{3 x} + c_{2} e^{x} - \frac{e ^{x} x ^{3}}{6} + \frac{e ^{x} x ^{2}}{2} + \frac{e ^{x} x}{2}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{3 x} + c_{2} e^{x} - \frac{e ^{x} x ^{3}}{6} + \frac{e ^{x} x ^{2}}{2} + \frac{e ^{x} x}{2}

y = c_{1} e^{3 x} + c_{2} e^{x} - \frac{e ^{x} x ^{3}}{6} + \frac{e ^{x} x ^{2}}{2} + \frac{e ^{x} x}{2}

What is the solution for y^{''}-4y^'+3y=(x^2-3x)e^x ?
The solution for y^{''}-4y^'+3y=(x^2-3x)e^x is y=c_{1}e^{3x}+c_{2}e^x-(e^xx^3)/6+(e^xx^2)/2+(e^xx)/2