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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

4 - 11 x + 6 x^{2}

x 3 + y 3 - 6 x y = 0

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

n = 0 \sum \infty sin (π \cdot n)

\frac{d}{d x} (6 x x)

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