What is the solution for 1+y^2+xyy^'=0 ?

The solution for 1+y^2+xyy^'=0 is y=(sqrt(c_{1)-x^2})/x ,y=-(sqrt(c_{1)-x^2})/x

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1+y^2+xyy^'=0

1 + y^{2} + x y y^{^{'}} = 0

y = \frac{c _{1} - x ^{2}}{x}, y = - \frac{c _{1} - x ^{2}}{x}

Solution steps

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

Solve separable ODE:

y = \frac{c _{1} - x ^{2}}{x}, y = - \frac{c _{1} - x ^{2}}{x}

y = \frac{c _{1} - x ^{2}}{x}, y = - \frac{c _{1} - x ^{2}}{x}

\frac{d y}{d x} = (1 + y^{2}) tan (x), y (0) = 3

3 (1 + t^{2}) d y = 2 t y (y^{3} - 1) d t

y^{^{'}} = y^{2} - 9

(x y + 2 y) d x + x d y = 0

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

What is the solution for 1+y^2+xyy^'=0 ?
The solution for 1+y^2+xyy^'=0 is y=(sqrt(c_{1)-x^2})/x ,y=-(sqrt(c_{1)-x^2})/x