The given equation:

\(\displaystyle{y}'=-\frac{{{y}^{{2}}+{x}^{{2}}}}{{{2}{x}{y}}},{y}{\left({1}\right)}={4}\)

The given ODE can be written as:

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

implies \(\displaystyle{y}'+{\left({\frac{{{1}}}{{{2}{x}}}}\right)}{y}={\left(-{\frac{{{x}}}{{{2}}}}\right)}{y}^{{-{{1}}}}\) (1)

A first oredr Bernoulli ODE has the form of

\(\displaystyle{y}'+{p}{\left({x}\right)}{y}={q}{\left({x}\right)}{y}^{{n}}\)

In this case we have also given a first order Bernoulli ODE with

\(\displaystyle{p}{\left({x}\right)}={\frac{{{1}}}{{{2}{x}}}},{q}{\left({x}\right)}=-{\frac{{{x}}}{{{2}}}},{n}=-{1}\)

Note that the general solution is obtained by substituting \(\displaystyle\upsilon={y}^{{{1}-{n}}}\) and solving \(\displaystyle{\frac{{{1}}}{{{1}-{n}}}}\upsilon'+{p}{\left({x}\right)}\upsilon={q}{\left({x}\right)}\)

Putting \(\displaystyle\upsilon={y}^{{2}}\) in (1) and using \(\displaystyle\upsilon'={2}{y}{y}'\) we get

\(\displaystyle\upsilon'+{\left(\frac{{1}}{{x}}\right)}=-{x}\)

Now by multipling the integrating factor \(\displaystyle{\exp{{\left(\int\frac{{1}}{{x}}{\left.{d}{x}\right.}\right)}}}={\exp{{\left({\ln{{x}}}\right)}}}={x}\) we get

\(\displaystyle\upsilon={\frac{{-{x}^{{3}}+{c}_{{1}}}}{{{3}{x}}}}\)

Now substitute back \(\displaystyle\upsilon={y}^{{2}}\)

\(\displaystyle{y}^{{2}}={\frac{{-{x}^{{3}}+{c}_{{1}}}}{{{3}{x}}}}\)

Apply initial conditions \(\displaystyle{y}{\left({1}\right)}={4}\) gives us \(\displaystyle{c}_{{1}}={49}\) and so we get \(\displaystyle{y}^{{2}}={\frac{{-{x}^{{3}}+{49}}}{{{3}{x}}}}\)

Note that \(\displaystyle{y}=-\sqrt{{{\frac{{-{x}^{{3}}+{49}}}{{{3}{x}}}}}}\), not possible as it does not gives us \(\displaystyle{y}{\left({1}\right)}={4}\)

So, the required solution is \(\displaystyle{y}=+\sqrt{{{\frac{{-{x}^{{3}}+{49}}}{{{3}{x}}}}}}\)