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

The solution for x^2y^{''}-xy^'+y= 1/x is y=c_{1}x+c_{2}ln(x)x+1/(4x)

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

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

y = c_{1} x + c_{2} ln (x) x + \frac{1}{4 x}

Solution steps

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

Solve linear ODE:

y = c_{1} x + c_{2} ln (x) x + \frac{1}{4 x}

y = c_{1} x + c_{2} ln (x) x + \frac{1}{4 x}

What is the solution for x^2y^{''}-xy^'+y= 1/x ?
The solution for x^2y^{''}-xy^'+y= 1/x is y=c_{1}x+c_{2}ln(x)x+1/(4x)