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?
 A: 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
$$
\int\frac{dy}{y\pm\sqrt{y^2+2}}.
$$
A: solving the quadratic gives
$$
y' = \frac12(y \pm \sqrt{y^2-1})
$$
which suggests the substitution $y=\cosh u$
A: We have
\begin{align*}
&y y^\prime = (y^\prime)^2 - 0.5 \\
& (y^\prime)^2 - y y^\prime = 0.5 \\
& (y^\prime)^2 - y y^\prime + \frac{1}{4}y^2 = 0.5 + \frac{1}{4} y^2 \\
& \left(y^\prime - \frac{1}{2} y \right)^2 = 0.5 + \frac{1}{4} y^2 \\
& y^\prime  = \frac{1}{2} \left(y  \pm \sqrt{1 + y^2} \right)
\end{align*}
You can solve this implicity using separation of variables.
A: If $y^\prime$ is large (compared to 1), neglect the -0.5 to get the approximate solution $y \sim \exp(\int dx\,f(x))$. If $y^\prime$ is small (compared to 1), neglect the ${y^\prime}^2$ on the RHS and solve.
Once you have an approximate solution, iterate. Or use it as a starting point to get a power series solution. A lot depends on what you know about $f(x)$ and about $y$ - a power series solution could be a good starting point if you have some idea of these two to begin with.
A: i will rewrite the differential equation $f(x)yy^\prime = (y^\prime)^2 - 0.5$ as a quadratic equation $$ 2\left(\frac{dy}{dx}\right)^2 - 2f(x)y\frac{dy}{dx} - 1 = 0.$$
this gives us $$\frac{dy}{dx} = {yf(x) \pm \sqrt{f^2(x)y^2 + 2} \over 2}$$
now i am stuck. i need to know a little more about $\ f.$ 
i will have to think more about this. but i will leave it here for a while. if i don't get any idea how to proceed i will delete it later. in the mean time hints are welcome.
