What is the solution for x^2y^{''}-xy^'+y-ln(x)=0 ?

The solution for x^2y^{''}-xy^'+y-ln(x)=0 is y=c_{1}x+c_{2}ln(x)x+ln(x)+2

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x^2y^{''}-xy^'+y-ln(x)=0

x^{2} y^{^{''}} - x y^{^{'}} + y - ln (x) = 0

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

Solution steps

x^{2} y^{''} (x) - x y^{'} (x) + y - ln (x) = 0

Solve linear ODE:

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

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

\frac{\partial}{\partial x} (4 cos (x) - e^{x^{2}})

\int_{1}^{2} \frac{4}{x ^{2}} d x

\int_{1}^{2} \frac{ln ( x )}{x} d x

\frac{d}{d x} (\frac{( 2 x ^{2} + 2 x + 2 )}{x})

\int \frac{1}{h x + a} d x

What is the solution for x^2y^{''}-xy^'+y-ln(x)=0 ?
The solution for x^2y^{''}-xy^'+y-ln(x)=0 is y=c_{1}x+c_{2}ln(x)x+ln(x)+2