What is the solution for (1+x^2)(1+y^2)dx-xydy=0 ?

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

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(1+x^2)(1+y^2)dx-xydy=0

(1 + x^{2}) (1 + y^{2}) d x - x y d y = 0

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

Solution steps

(1 + x^{2}) (1 + y^{2}) d x - x y d y = 0

Solve separable ODE:

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

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

\int (6 sin (3 x) - cos (2 x)) d x

\int x - 4 d x

x \to 0 + lim (\frac{3}{x ^{2} - x})

\int \frac{16 s + 16}{( s ^{2} + 1 ) ( s - 1 ) ^{3}} d s

x \to 0.999 lim (x^{2} + 2)

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