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

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

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

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

y = c_{1} e^{3 x} + c_{2} e^{x} - \frac{e ^{x} x ^{3}}{6} - \frac{e ^{x} x ^{2}}{4} - \frac{e ^{x} x}{4}

Solution steps

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

y = c_{1} e^{3 x} + c_{2} e^{x} - \frac{e ^{x} x ^{3}}{6} - \frac{e ^{x} x ^{2}}{4} - \frac{e ^{x} x}{4}

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

y^{^{''}} + 9 y = 2 csc^{2} (3 x)

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

y^{^{''}} + 4 y^{^{'}} + 3 y = 1, y^{^{'}} (0) = 1

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

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