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^{5/3})/(10)

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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{5}{3}}}{10}

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{5}{3}}}{10}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{9 e ^{x} x ^{\frac{5}{3}}}{10}

y^{^{''}} + a^{2} y^{^{'}} = cos (b x)

y^{^{''}} - 5 y^{^{'}} + 4 y = 2 x^{2}

y^{^{''}} + 16 y = 7 x, y (0) = 1, y^{^{'}} (0) = 4

y^{^{''}} - 4 y^{^{'}} + 3 y = 10 cos (x)

y^{^{''}} + 4 y = 3 x (2 + cos (2 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^{5/3})/(10)