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}}

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

y^{^{''}} - 5 y^{^{'}} + 6 e^{x} = 0

3 \frac{d ^{2} y}{d x ^{2}} - 4 \frac{d y}{d x} + y = x - 7

y^{^{''}} - 5 y^{^{'}} + 6 y = 10 sin (2 x) - 50 cos (2 x)

y^{^{''}} - 10 y^{^{'}} + 25 y = t^{- 4} e^{5 t}

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}