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

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

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

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

y = \frac{2 x}{- ln ( x ) + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{2 x}{- ln ( x ) + c _{1}}

y = \frac{2 x}{- ln ( x ) + c _{1}}

\frac{d}{d x} (\frac{( x - 4 )}{x} + ln (\frac{x}{4}))

\frac{\partial}{\partial y} (\frac{x}{2 y ^{4}})

3 x + 1

\int \frac{1}{x ^{4} + x ^{2}} d x

x^{2} e^{\frac{1}{x}}

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