Chapter 2, miscellaneous exercises no. 4:

$\displaystyle x y' = \sqrt {x^2 + y^2}$

I made it homogeneous:

$\displaystyle \frac {dy} {dx} = \sqrt {1 + \left({\frac y x}\right)^2}$

Then I substituted $\displaystyle z = \frac x y$ which makes it separable:

$\displaystyle z + x \frac {dz}{dx} = \sqrt {1 + z^2}$

and so

$\displaystyle \int \frac {dz} {\sqrt {1 + z^2} - z} = \int \frac {dx} x$

Now the nightmare begins. The integral on the LHS seems to be tractable only if I try the substitution $\displaystyle z = \sinh w$

This leads to:

$\displaystyle \int \frac {\cosh w dw} {\cosh w - \sinh w}$

Converting the hyp funcs to exps:

$\displaystyle \int \frac {e^w + e^{-w}} {e^{-2w}}$

which gets us to

$\displaystyle \int \left({e^{3w} + e^w}\right) dw = \frac {e^{3w}} 3 + e^w + C$

To get back to $\displaystyle z$ and thence $\displaystyle y/x$, I substitute back using the formula for $\displaystyle w = \sinh^{-1} z = \ln \left({z + \sqrt{z^2 + 1}}\right)$ from which I end up with the answer to my diffeq:

$\displaystyle \left({\frac y x + \sqrt{\left({\frac y x}\right)^2 + 1}}\right)^3 + 3 \left({\frac y x + \sqrt{\left({\frac y x}\right)^2 + 1}}\right) = 3 \ln x + C$

So far so good, but the answer as given in the book is:

$\displaystyle y \sqrt {x^2 + y^2} + x^2 \ln \left({y + \sqrt {x^2 + y^2}}\right) + y^2 = 3 x^2 \ln x + c x^2$

Can these two results be reconciled? I fear not. Can anyone work it through and see whether (a) I've made a mistake somewhere, or (b) I haven't followed through far enough?