# Basic differential equations, homogeneous method

1. Jul 7, 2008

### 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

Last edited: Jul 7, 2008
2. Jul 7, 2008

### 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?