# Math Help - [SOLVED] 1st order homogeneous diffeq from G.F.Simmons

1. ## [SOLVED] 1st order homogeneous diffeq from G.F.Simmons

Chapter 2, miscellaneous exercises no. 4:

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

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

Then I substituted $z = \frac x y$ which makes it separable:
$z + x \frac {dz}{dx} = \sqrt {1 + z^2}$

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

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

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

Converting the hyp funcs to exps:
$\int \frac {e^w + e^{-w}} {e^{-2w}}$

which gets us to
$\int \left({e^{3w} + e^w}\right) dw = \frac {e^{3w}} 3 + e^w + C$

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

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

So far so good, but the answer as given in the book is:
$y \sqrt {x^2 + y^2} + x^2 \ln \left({y + \sqrt {x^2 + y^2}}\right) + y^2 = 3 x^2 \ln x + c x^2$

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?

2. Originally Posted by Matt Westwood
[snip]

Chapter 2, miscellaneous exercises no. 4:

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

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

Then I substituted $z = \frac x y$ which makes it separable:
$z + x \frac {dz}{dx} = \sqrt {1 + z^2}$

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

[/snip]
Here's the trick (for LHS):

$\int\frac{\,dz}{\sqrt{1+z^2}-z}=\int\frac{1}{\sqrt{1+z^2}-z}\cdot\frac{\sqrt{1+z^2}+z}{\sqrt{1+z^2}+z}\,dz$ $=\int\frac{\sqrt{1+z^2}+z}{1+z^2-z^2}\,dz=\int\sqrt{1+z^2}+z\,dz$.

Now all you need to do is apply a trig sub to $\int\sqrt{1+z^2}\,dz$; it should be a walk in the park from there.

3. Ahh, gotcha. Genius. I'd completely forgotten about that sweet little manipulation.

I like your DE tutorial, btw.