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}}}

\int_{1}^{4} \frac{5}{x} d x

\frac{d}{d x} (\frac{x}{2} - ln (3))

\frac{d x}{d t} = - a x

\int_{3}^{7} \frac{x}{x ^{2} - 2} d x

y = x^{2} - 5 x + 5 (0.5)

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}}