1. ## First Order Differential

$2r(s^2+1)dr+(r^4+1)ds=0$

This is the second question in the textbook and I am able to do every question after it, so perhaps there is something simple that I am not seeing? The answer has no trig functions in it.

$(2r)/(r^4+1)dr+(s^2+1)ds=0$

The only way I can integrate the dr term is with trig functions, but according the the answers there is a simple way such that no trig functions are required.

2. ## Re: First Order Differential

Originally Posted by quantoembryo
$2r(s^2+1)dr+(r^4+1)ds=0$

This is the second question in the textbook and I am able to do every question after it, so perhaps there is something simple that I am not seeing? The answer has no trig functions in it.

$(2r)/(r^4+1)dr+(s^2+1)ds=0$

The only way I can integrate the dr term is with trig functions, but according the the answers there is a simple way such that no trig functions are required.
Hi quantoembryo!

After separation of variables, your equation should be:

$\frac{2r}{r^4+1}dr+\frac{1}{s^2+1}ds=0$

How would you integrate the ds term?
Can you perhaps apply something similar to the dr term?

3. ## Re: First Order Differential

The first term can be integrated with a substitution if you write \displaystyle \begin{align*} \frac{2r}{r^4 + 1} = \frac{2r}{\left( r^2 \right)^2 + 1} \end{align*} and let \displaystyle \begin{align*} u = r^2 \implies du = 2r\,dr \end{align*}.

4. ## Re: First Order Differential

Ah. So simple. Thanks!