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}

cos^{4} (x)

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

2 x^{2} - 1, 0, 1

1 - 4 x^{2}

\frac{x + 5}{x}

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)