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

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

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

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

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

Solution steps

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

Solve Bernoulli ODE:

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

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

cos (x) y^{^{'}} + sin (x) y = 0

x^{2} + 24 x + c

\frac{d}{d x} (100 x)

\int \frac{2 x ^{3}}{x ^{2} + 1} d x

x \to 0 lim (x \cdot ln (x))

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