Basic differential equations, homogeneous method

[MAins]

xy'-y =sqrt((x^2)+y(^2))
so by method of homogeneous equations I get arcsinh(y/x) = ln|x|+c but don't know what to do from there. the book suggests using sinhx = 1/2 (e^x - e^-x) but I don't know how to incorporate this. The answer should be y= (1/2) * (cx^2 - 1/c). How do you get this?

This is my work so far:
y'-(y/x) = sqrt (1+(y/x)^2)
since y=ux...
du/(sqrt(1+u^2) = dx/x
taking the antiderivatives gives
arcsinh(u) = ln|x|+c
where
arcsinh|y/x| = ln|x|+c

 

[rock.freak667]

$sinh^{-1}(\frac{y}{x}) = lnx + ln A$ (I put c=lnA to make things simpler)
$\Rightarrow sinh^{-1}(\frac{y}{x}) = lnAx$


$$\Rightarrow \frac{y}{x} = sinh(lnAx)$$


and they said that


$$sinhX=\frac{e^X - e^{-X}}{2}$$


So $sinh(lnAx)$ should make the equation

$$\frac{y}{x}=\frac{e^{lnAx}-e^{-lnAx)}}{2}$$

Do you see how to continue from here?