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

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

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y^'xy=x^2+2y^2

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

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

Solution steps

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

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}

\frac{d}{d x} ((\frac{1 - 2 x}{1 + 3 x})^{4})

\int_{- 9}^{9} 1 + x^{2} d x

\frac{2}{1 + x ^{2}}

\int \frac{1}{x ^{2} + 12 x + 85} d x

y = \frac{1}{2} x^{3} + 2, y = - x + 1, x = 0, x = 2

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