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

The solution for ln(x)-x^2y=xy^' is y=(integral of (ln(x)e^{\frac{x^2)/2})/(x)}{e^{(x^2)/2}}

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ln(x)-x^2y=xy^'

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

y = \frac{\int \frac{l n ( x ) e ^{\frac{x ^{2}}{2}}}{x} d x}{e ^{\frac{x ^{2}}{2}}}

Solution steps

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

Solve linear ODE:

y = \frac{\int \frac{l n ( x ) e ^{\frac{x ^{2}}{2}}}{x} d x}{e ^{\frac{x ^{2}}{2}}}

y = \frac{\int \frac{l n ( x ) e ^{\frac{x ^{2}}{2}}}{x} d x}{e ^{\frac{x ^{2}}{2}}}

What is the solution for ln(x)-x^2y=xy^' ?
The solution for ln(x)-x^2y=xy^' is y=(integral of (ln(x)e^{\frac{x^2)/2})/(x)}{e^{(x^2)/2}}