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

The solution for y^{''}-2y^'+y=x^3e^x is y=c_{1}e^x+c_{2}xe^x+(e^xx^5)/(20)

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{5}}{20}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{5}}{20}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{5}}{20}

y^{^{''}} + 2 y - 1 = 0

y^{^{''}} + 5 y^{^{'}} + 6 y = 6 t + 17, y (0) = 3, y^{^{'}} (0) = - 4

y^{^{''}} - 3 y = e^{x} sin (5 x)

2 y^{^{''}} - 4 y^{^{'}} + 10 y = 2, x y (0) = 0, y^{^{'}} (0) = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = t - 2, y (0) = 0, y^{^{'}} (0) = 1

What is the solution for y^{''}-2y^'+y=x^3e^x ?
The solution for y^{''}-2y^'+y=x^3e^x is y=c_{1}e^x+c_{2}xe^x+(e^xx^5)/(20)