What is the solution for (4x^2-y^2)dx-xydy=0 ?

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

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(4x^2-y^2)dx-xydy=0

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

y = \frac{2 x ^{4} + c _{1}}{x}, y = - \frac{2 x ^{4} + c _{1}}{x}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{2 x ^{4} + c _{1}}{x}, y = - \frac{2 x ^{4} + c _{1}}{x}

y = \frac{2 x ^{4} + c _{1}}{x}, y = - \frac{2 x ^{4} + c _{1}}{x}

y^{^{'}} = 2 (2 x + y)^{2}

\frac{d}{d x} (13 x e^{- x})

\frac{\partial}{\partial v} (\frac{1}{4} (u - v))

f (x) = (x - 1)^{2}

n = 1 \sum \infty e^{\frac{1}{n}}

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