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}

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

\int (x^{2} + 4)^{2} d x

x \to \infty lim (- 5 x^{3})

\int x^{3} - 3 x d x

\frac{d}{d x} (100 e^{x})

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)}