What is the solution for (dy)/(dx)=(1-xy^2)/(2x^2y) ?

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

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(dy)/(dx)=(1-xy^2)/(2x^2y)

\frac{d y}{d x} = \frac{1 - x y ^{2}}{2 x ^{2} y}

y = \frac{ln ( x ) + c _{1}}{x}, y = - \frac{ln ( x ) + c _{1}}{x}

Solution steps

\frac{d y}{d x} = \frac{1 - x y ^{2}}{2 x ^{2} y}

Solve Bernoulli ODE:

y = \frac{ln ( x ) + c _{1}}{x}, y = - \frac{ln ( x ) + c _{1}}{x}

y = \frac{ln ( x ) + c _{1}}{x}, y = - \frac{ln ( x ) + c _{1}}{x}

\frac{\partial}{\partial x} (z (2 x y))

y^{^{'}} + y - 2 = 0

x \to 2 lim (2 x^{2} - 7 x + 4)

\int tan^{3} (θ) d θ

\frac{d}{d x} (e^{s e c (l n (x^{2} + 1))})

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