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+(4e^xx^{5/2})/(15)

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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 e ^{x} x ^{\frac{5}{2}}}{15}

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 e ^{x} x ^{\frac{5}{2}}}{15}

y = c_{1} e^{x} + c_{2} x e^{x} + \frac{4 e ^{x} x ^{\frac{5}{2}}}{15}

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

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

y^{^{''}} - 6 y^{^{'}} + 9 y = t^{2} e^{3 t}

y^{^{''}} - 25 y = cosh (5 x)

y^{^{''}} - 3 y^{^{'}} - 4 y = e^{- x}

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+(4e^xx^{5/2})/(15)