How Do You Solve This Differential Equation Involving Trigonometric Identities?

[Arman777]

Homework Statement

$2r(s^2+1)dr+(r^4+1)ds=0$
Find the solution of the given Diff. Eqn.

Homework Equations

s[/B]

The Attempt at a Solution

its a separable equation.

Hence I divided both sides with $\frac {1} {(s^2+1)(r^4+1)}$ and we get
$\frac {2r} {(r^4+1)} dr+\frac {1} {(s^2+1)}ds=0$
since its in the form of $g(r)dr+h(s)ds=0$
I can take normal integral and I get

$arctan(r^2)+arctan(s)=arctan(c)$

but from now on I didnt know what to do

I thought to write $r^2+s=c$ but answer is
$r^2+s=c(1-r^2s)$ which I didnt understand where the $(1-r^2s)$ comes from ?

 

[Ray Vickson]

Homework Statement

$2r(s^2+1)dr+(r^4+1)ds=0$
Find the solution of the given Diff. Eqn.

Homework Equations

s[/B]

The Attempt at a Solution

its a separable equation.

Hence I divided both sides with $\frac {1} {(s^2+1)(r^4+1)}$ and we get
$\frac {2r} {(r^4+1)} dr+\frac {1} {(s^2+1)}ds=0$
since its in the form of $g(r)dr+h(s)ds=0$
I can take normal integral and I get

$arctan(r^2)+arctan(s)=arctan(c)$

but from now on I didnt know what to do

I thought to write $r^2+s=c$ but answer is
$r^2+s=c(1-r^2s)$ which I didnt understand where the $(1-r^2s)$ comes from ?

The identity
$$\tan(a+b) = \frac{\tan a + \tan b}{1- \tan a \: \tan b}$$
implies that
$$\arctan(u) + \arctan(v) = \arctan \left( \frac{u+v}{1-uv} \right)$$

 

[Orodruin]

I suggest that you look up the trigonometric identity for $\tan(a+b)$ expressed in $\tan(a)$ and $\tan(b)$ (or even better, derive it yourself if you are able).

 

[Arman777]

Hmm, I see. I never thought that actually :/ Thanks :)

 

[Arman777]

I suggest that you look up the trigonometric identity for $\tan(a+b)$ expressed in $\tan(a)$ and $\tan(b)$ (or even better, derive it yourself if you are able).

I'll try to do