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

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

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

(4 x^{2} + 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

(4 x^{2} + 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}

\frac{d y}{d x} = x - 2 y + 12

f (x) = 5 x + 7

f (x) = \frac{5}{x}, at P (4, \frac{5}{4})

y^{^{'}} = 8 x y - 2 x

\int - 15 x^{\frac{2}{3}} d x

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