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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

\int \frac{1}{- x ^{2} + 4 x - 3} d x

\int \frac{15}{4} x^{2} d x

\frac{d}{d t} (lo g_{3} (2 t - x))

y = 2 x^{3} + 3 x^{2} - 12 x + 1

\frac{d}{d r} (r + 3 r)

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