What is the solution for (x^2-3y^2)dx+(2xy)dy=0 ?

The solution for (x^2-3y^2)dx+(2xy)dy=0 is y=xsqrt(1+c_{1)x},y=-xsqrt(1+c_{1)x}

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(x^2-3y^2)dx+(2xy)dy=0

(x^{2} - 3 y^{2}) d x + (2 x y) d y = 0

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

Solution steps

(x^{2} - 3 y^{2}) d x + (2 x y) d y = 0

Solve Bernoulli ODE:

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

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

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

f (x) = 4 x^{2} - 4 x + 1, (- 1, 8)

\frac{\partial}{\partial x} (y - \frac{27}{x ^{2}})

\int (2 sin (x) - 5 e^{x}) d x

\int \frac{1}{2} sin (2 θ) d θ

What is the solution for (x^2-3y^2)dx+(2xy)dy=0 ?
The solution for (x^2-3y^2)dx+(2xy)dy=0 is y=xsqrt(1+c_{1)x},y=-xsqrt(1+c_{1)x}