What is the solution for (y^2+yx)dx+(-x^2)dy=0 ?

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

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(y^2+yx)dx+(-x^2)dy=0

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

y^{^{'}} = \frac{y ^{2} + 4 x x ^{2} + y ^{2}}{x y}

ln (4 x + 1)

3 x^{2} - 4 x + 1

\frac{d}{d x} ((8 (3 x)^{4} + 5)^{3})

\int \frac{20 x ^{2} + x + 45}{x ( 4 x ^{2} + 9 )} d x

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