<s>[INST] Solving the differential equation: $f(x)yy'=(y')^2-0.5$ I am trying to solve this equation:
$f(x)yy'=(y')^2-0.5$
I have already tried traditional methods...
Any ideas?
 [/INST] We have
$$
\left(\frac{y'}{y}\right)^2-\frac{y'}{y}=\frac{1}{2y^2},
$$
or
$$
\left(\frac{y'}{y}-\frac{1}{2}\right)^2=\frac{1}{4}+\frac{1}{2y^2},
$$
or
$$
\frac{y'}{y}=\frac{1}{2}\pm\sqrt{\frac{1}{4}+\frac{1}{2y^2}}
$$
or
$$
y'=\frac{y}{2}\pm\sqrt{\frac{y^2}{4}+\frac{1}{2}}=\frac{1}{2}\big(y\pm\sqrt{y^2+2}\big).
$$
Using now Separation of Variables the problem reduces to finding the integral
$$
\int\frac{dy}{y\pm\sqrt{y^2+2}}.
$$
</s>