What is the solution for y^'=(x^2-y^2)/(xy),y(2)=4 ?

The solution for y^'=(x^2-y^2)/(xy),y(2)=4 is y=(sqrt(x^4+112))/(sqrt(2)x)

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y^'=(x^2-y^2)/(xy),y(2)=4

y^{^{'}} = \frac{x ^{2} - y ^{2}}{x y}, y (2) = 4

y = \frac{x ^{4} + 112}{2 x}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x ^{4} + 112}{2 x}

y = \frac{x ^{4} + 112}{2 x}

12 x sin (x)

x \to 2 lim ((2 x - 1)^{2} - x^{2})

\int_{0}^{\infty} x e^{- \frac{x}{4}} d x

y = x^{2}, (- 0.7, 0.49)

\int (x - 8) sin (π x) d x

What is the solution for y^'=(x^2-y^2)/(xy),y(2)=4 ?
The solution for y^'=(x^2-y^2)/(xy),y(2)=4 is y=(sqrt(x^4+112))/(sqrt(2)x)