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

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

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

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

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

Solution steps

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

Solve linear ODE:

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

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

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