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}}

x \to 2 lim (x^{2} + 8)

x^{7} e^{x} 5

x \to 0 lim (\frac{sin ^{2} ( x )}{x})

y = 1 x^{2}

x \to - 1 lim (e^{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)}