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

The solution for y^'=(x^2-y^2)/(xy),y(2)=2 is y=(sqrt(x^4+16))/(sqrt(2)x)

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

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

y = \frac{x ^{4} + 16}{2 x}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x ^{4} + 16}{2 x}

y = \frac{x ^{4} + 16}{2 x}

4^{6 x + 7}

\int 3 sin (x) cos (x) d x

y^{^{''}} + 9 y = \frac{1}{sin ( 3 t )}

y = 2 (x + 1), y = 3 (x + 1), x = 7

\frac{( t + 5 ) 2}{3 ( 2 t ^{2} - 3 ) 3}

What is the solution for y^'=(x^2-y^2)/(xy),y(2)=2 ?
The solution for y^'=(x^2-y^2)/(xy),y(2)=2 is y=(sqrt(x^4+16))/(sqrt(2)x)