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}{d x} (- 2 x^{5} + \frac{1}{2 x ^{2}})

y^{^{''}} + 25 y = - 50 sin (5 t)

\frac{d}{d x} (1 - (1 + \frac{x}{200}) e^{- \frac{x}{200}})

\int_{0}^{3 π} sin (x) d x

\frac{1}{5 + 7 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)}