# First Order Differential

• Mar 11th 2013, 01:49 PM
quantoembryo
First Order Differential
$\displaystyle 2r(s^2+1)dr+(r^4+1)ds=0$

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.

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

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.
• Mar 11th 2013, 01:58 PM
ILikeSerena
Re: First Order Differential
Quote:

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

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.

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

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:

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

How would you integrate the ds term?
Can you perhaps apply something similar to the dr term?
• Mar 11th 2013, 02:17 PM
Prove It
Re: First Order Differential
The first term can be integrated with a substitution if you write \displaystyle \displaystyle \begin{align*} \frac{2r}{r^4 + 1} = \frac{2r}{\left( r^2 \right)^2 + 1} \end{align*} and let \displaystyle \displaystyle \begin{align*} u = r^2 \implies du = 2r\,dr \end{align*}.
• Mar 11th 2013, 02:51 PM
quantoembryo
Re: First Order Differential
Ah. So simple. Thanks!