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}

\int e^{6 u} d u

f (x) = \frac{x - 9}{8 x - 1}, at x = 1

lo g_{3} (x)

y^{^{'''}} + 6 y^{^{''}} - 15 y^{^{'}} + 8 y = 0

\frac{d}{d x} (\frac{x ^{2}}{9 + 4 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)