What is the solution for 2xyy^'=y^2-x^2 ?

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

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2xyy^'=y^2-x^2

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

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

Solution steps

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

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

h \to 0 lim (- 2 x - h + 4)

\int z z^{2} - 5 d z

cos^{2} (5 t)

\int_{0}^{\infty} \frac{3}{x ^{2} + 4} d x

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

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