What is the solution for y^'=-(2x)/y ?

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

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y^'=-(2x)/y

y^{^{'}} = - \frac{2 x}{y}

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

Solution steps

y^{'} (x) = - \frac{2 x}{y}

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 \to 1 lim (∣ e^{x} - 1 ∣)

\int \frac{2}{3 x} d x

\frac{d y}{d x} = \frac{x}{y}, y (0) = - 7

y^{^{''}} + \frac{3}{x} y^{^{'}} = 0

\frac{d y}{d x} = \frac{8 x}{5 y}

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