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

The solution for 2(2x^2+y^2)dx-xydy=0 is y=xsqrt(-4+c_{1)x^2},y=-xsqrt(-4+c_{1)x^2}

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2(2x^2+y^2)dx-xydy=0

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

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

Solution steps

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

Solve Bernoulli ODE:

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

y = x - 4 + c_{1} x^{2}, y = - x - 4 + c_{1} x^{2}

f (x) = \frac{x - 5}{x + 3}

f (x) = 5 x^{2} - 2 x + 4

x \to - 4 + lim (\frac{- 1}{x ^{2} - 16})

x \to 1 lim (\frac{2 x - 7}{4 x + 5})

\frac{2}{s - 2}

What is the solution for 2(2x^2+y^2)dx-xydy=0 ?
The solution for 2(2x^2+y^2)dx-xydy=0 is y=xsqrt(-4+c_{1)x^2},y=-xsqrt(-4+c_{1)x^2}