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

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

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(dy)/(dx)=(x^2-y^2)/(xy),y(2)=2

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

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

Solution steps

\frac{d y}{d 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 x, x^{2} - 5

\int se x (x)

f (x) = x^{2} - 4, 1, 3

y = \frac{7}{x}

\int_{0}^{\infty} 12 e^{- 6 x} d x

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