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

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

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y^'=(x^2+2y^2)/(xy)

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

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