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

The solution for y^{''}-2y^'+y=(x^2-2x)e^x is y=c_{1}e^x+c_{2}xe^x+(e^xx^4}{12}-\frac{e^xx^3)/3

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

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{4}}{12} - \frac{e ^{x} x ^{3}}{3}

Solution steps

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

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{4}}{12} - \frac{e ^{x} x ^{3}}{3}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{e ^{x} x ^{4}}{12} - \frac{e ^{x} x ^{3}}{3}

y^{^{''}} + y^{^{'}} - 2 y = t e^{3 t}

y^{^{''}} + 24 y^{^{'}} + 144 y = t^{2} e^{- 12 t} + e^{- 12 t}

y^{^{''}} - 81 y = 3

y^{^{''}} - 5 y^{^{'}} + 4 y = e^{2 t}, y (0) = 1, y^{^{'}} (0) = - 1

y^{^{''}} + 4 y = 4 sin (2 t), y (π) = 0, y^{^{'}} (π) = - 1

What is the solution for y^{''}-2y^'+y=(x^2-2x)e^x ?
The solution for y^{''}-2y^'+y=(x^2-2x)e^x is y=c_{1}e^x+c_{2}xe^x+(e^xx^4}{12}-\frac{e^xx^3)/3