What is the solution for 2xydx+y^2dy=0 ?

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

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2xydx+y^2dy=0

2 x y d x + y^{2} d y = 0

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

Solution steps

2 x y d x + y^{2} d y = 0

Solve separable ODE:

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

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

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

\frac{d v}{d t} = g - k v (t)

\frac{d z}{d t} + 3 e^{t + z} = 0

2 x^{6} y^{^{'}} + 9 x^{8} y^{4} = 0

(1 + x) y^{^{'}} = 4 y, y (0) = 2

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