What is the solution for y^'=((x^2-y^2))/(xy),y(2)=3 ?

The solution for y^'=((x^2-y^2))/(xy),y(2)=3 is y=(sqrt(x^4+56))/(sqrt(2)x)

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y^'=((x^2-y^2))/(xy),y(2)=3

y^{^{'}} = \frac{( x ^{2} - y ^{2} )}{x y}, y (2) = 3

y = \frac{x ^{4} + 56}{2 x}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{x ^{4} + 56}{2 x}

y = \frac{x ^{4} + 56}{2 x}

y = sin (π - x)

f (x) = 7 x - 17, at x = 3

\frac{d}{d x} (6 x^{\frac{1}{3}} + 3 x^{\frac{4}{3}})

\int \frac{x + 1}{x} d x

(x^{2} - 2 x + 3)^{^{'}}

What is the solution for y^'=((x^2-y^2))/(xy),y(2)=3 ?
The solution for y^'=((x^2-y^2))/(xy),y(2)=3 is y=(sqrt(x^4+56))/(sqrt(2)x)