What is the solution for (y^2-xy)dx+x^2dy=0,μ= 1/(x^2) ?

The solution for (y^2-xy)dx+x^2dy=0,μ= 1/(x^2) is y= x/(ln(x)+c_{1)}

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(y^2-xy)dx+x^2dy=0,μ= 1/(x^2)

(y^{2} - x y) d x + x^{2} d y = 0, μ = \frac{1}{x ^{2}}

y = \frac{x}{ln ( x ) + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x}{ln ( x ) + c _{1}}

y = \frac{x}{ln ( x ) + c _{1}}

(2.1) (1.6)

ln (\frac{x ^{2}}{3 x + 1})

(x + 1) y^{^{'}} = x + y

\int \frac{x}{9 + x ^{2}} d x

\frac{d}{d x} (\frac{6 cos ( x )}{1 - sin ( x )})

What is the solution for (y^2-xy)dx+x^2dy=0,μ= 1/(x^2) ?
The solution for (y^2-xy)dx+x^2dy=0,μ= 1/(x^2) is y= x/(ln(x)+c_{1)}