What is the solution for 1/(y^')=(2y)/x ?

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

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1/(y^')=(2y)/x

\frac{1}{y ^{^{'}}} = \frac{2 y}{x}

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

Solution steps

\frac{1}{y ^{'} ( x )} = \frac{2 y}{x}

Solve separable ODE:

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

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

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