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

The solution for y^{''}-2y^'+y=x^2e^x is y=c_{1}e^x+c_{2}xe^x+(e^xx^4)/(12)

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{4}}{12}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{4}}{12}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{4}}{12}

(d^{2} - 4 d - 12) \cdot y = (x - 1) \cdot e^{2 x}

(d^{2} + 3 d + 2) \cdot y = x^{2} + sin (x)

(d^{2} - 6 d + 9) y = 2 e^{3 x}

(d^{2} - 1) y = sech (x)

(d + 2) \cdot (d - 1)^{3} \cdot y = e^{x}

What is the solution for y^{''}-2y^'+y=x^2e^x ?
The solution for y^{''}-2y^'+y=x^2e^x is y=c_{1}e^x+c_{2}xe^x+(e^xx^4)/(12)