A solution to the differential equation $x^2y'+xy=\sqrt{x^2y^2+1}$ I need help to start solving a differential equation
$$x^2y'+xy=\sqrt{x^2y^2+1}.$$
I would divide the equation with $x^2.$ Then the equation looks like a homogeneous equation, but I get under the square root $\frac{1}{x^4}.$  I don't know what else to do. Can someone give me a hint?
 A: let $z=xy$
$y'=\frac{xz'-z}{x^{2}}$
$xz'=\sqrt{{z^{2}}+1}$
$\displaystyle \int \frac{1}{\sqrt{z^2+1}} \,dz = \displaystyle \int \frac{1}{x} \,dx$
$\operatorname{arcsinh}(z)=\ln(x)+C$
$z=\sinh(\ln(x)+c)$
$$y=\frac{\sinh(\ln(x)+c)}{x}$$
A: Set some variable, lets call him q, as $$q = xy$$
then, $$\frac{dy}{dx} = \frac{xq'-q}{x^{2}}$$
$$xq' = \sqrt{q^{2}+1}$$
$$\displaystyle \int \frac{1}{\sqrt{q^2+1}} \,dq = \displaystyle \int \frac{1}{x} \,dx$$
$$arcsinh(q) = \ln(x) + C$$
$$q = \sinh(\ln(x) + C)$$
Hope it helped.
A: But this is not the only answer to the equation!!!
$\operatorname{arcsinh}(z)=\ln(x)+\ln(c)$
$\ln(\sqrt{z^{2}+1}+z)=\ln(cx)$
$\sqrt{z^{2}+1}+z=cx$
$z^{2}+1=z^{2}-2cxz+(cx)^{2}$
$z=\frac{cx}{2}-\frac{1}{2cx}$
$$y=\frac{c}{2}-\frac{1}{2cx^{2}}$$
This form is another general answer. Now, for the abnormal solution, we derive the general solution with respect to c
$0=\frac{1}{2}+\frac{1}{2c^{2}x^{2}}$
\begin{cases} c=\frac{\operatorname i}{x} \\\\
c=\frac{-\operatorname i}{x}\\\\
\end{cases}
so
\begin{cases} y=\frac{\operatorname i}{x} \\\\
y=\frac{-\operatorname i}{x}\\\\
\end{cases}
