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}

a e^{- b^{2} x^{2}}

e^{2}

\frac{\partial}{\partial x} (3 x^{2} y^{2})

3 t^{2} + 3 t + 1

f (x) = x^{6} \cdot e^{x}

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}