I was given the following ode $y'=\frac{y}{x}+tan(\frac{y}{x})$ I found the answer which is: $y=arcsin(kx)x$
Now I need to find the singular solution, if I look at the ode $y'=\frac{y}{x}+tan(\frac{y}{x})$ if $x=0$ then the function is "undefined" that what I need to look when searching for the singular solution? values in which the ODE is not defined?
HINT : $$y'=\frac{y}{x}+\tan(\frac{y}{x})$$
This is an homogeneous ODE (Second meaning of homogeneous, i.e. involving $\frac{y}{x}$). The usual method of solving is the change of function : $y(x)=xf(x)$ .
$$f+xf'=f+\tan(f) \quad\to\quad xf'=\tan(f)$$
The original ODE is changed to a separable ODE easy to solve, which finally leads to $$y=x\sin^{-1}(c\:x)$$ This is the general solution that you already obtained.
The case $c=0$ is singular since $\quad\sin^{-1}(0\:x)=n\pi\:\quad$ which isn't function of $x$. This corresponds to the singular solutions : $$y=n\pi\:x$$ $n$ is any integer.