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}

\int 4 x - 6 d x

\int \frac{8 x - 3}{12 x ^{2} - 7 x + 1} d x

\int \frac{x ^{3}}{x ^{4} + 4} d x

(\frac{1}{y ^{2}} - \frac{9}{y ^{4}}) (y + 5 y^{3})

f (x) = ln (\frac{( 1 + x )}{( 1 - 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)}