What is the solution for y^{''}-2y^'+y=x^{1/3}e^x ?

The solution for y^{''}-2y^'+y=x^{1/3}e^x is y=c_{1}e^x+c_{2}xe^x+(9e^xx^{7/3})/(28)

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y^{''}-2y^'+y=x^{1/3}e^x

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

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{9 e ^{x} x ^{\frac{7}{3}}}{28}

Solution steps

y^{''} (x) - 2 y^{'} (x) + y = x^{\frac{1}{3}} e^{x}

Solve linear ODE:

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{9 e ^{x} x ^{\frac{7}{3}}}{28}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{9 e ^{x} x ^{\frac{7}{3}}}{28}

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

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

y^{^{''}} + 4 y = 9 t

y^{^{''}} + 4 y = 3 csc (x)

y^{^{'''}} - 6 y^{^{''}} + 11 y^{^{'}} - 6 y = 2 x e^{- x}

What is the solution for y^{''}-2y^'+y=x^{1/3}e^x ?
The solution for y^{''}-2y^'+y=x^{1/3}e^x is y=c_{1}e^x+c_{2}xe^x+(9e^xx^{7/3})/(28)