
First order ODEs
Greetings, I have a few differential equation questions and I began to take this course at the beginning of this semester so I might have been doing obvious mistakes.These questions are from Ross, Differential Equations 1st edition chapter 2.2.
3.$\displaystyle 2r(s^2+1)dr+(r^4+1)ds=0$
14.$\displaystyle (\sqrt(x+y)+\sqrt(xy))dx+(\sqrt(xy)\sqrt(x+y))dy=0$
23.(a)Prove that if $\displaystyle Mdx+Ndy=0$ is a homogeneous equation, then the change of variables $\displaystyle x=uy$ transforms this equation into a separable equation in the variables u and x.
(b)Use this result to solve $\displaystyle (x^23y^2)dx+2xydy=0$
(c)Use this result to solve $\displaystyle (y+\sqrt(x^2+y^2))dxxdy=0$
In Q3 I found $\displaystyle \arctan(r^2)+\arctan(s)=C$ and unable to reduce this expresion to $\displaystyle r^2+s=c(1r^2s)$ as given in the answers at the back of the book. In Q14 I got the following expression and unable to proceed further:
$\displaystyle \int(\frac{(\sqrt(1+v)\sqrt(1v))dv}{\sqrt(1+v)+\sqrt(1v)+v\sqrt(1v)v\sqrt(1+v)})=\lnx+C$ where $\displaystyle v=\frac{y}{x}$.
In Q23 I started with $\displaystyle \frac{dy}{dx}=h(x/y)$ then letting x=uy leads $\displaystyle \frac{dy}{dx}=\frac{1}{u}\frac{x}{u^2}\frac{du}{dx}=h(u)$
It follows that:
$\displaystyle \int\frac{dx}{x}\int\frac{du}{u^2(\frac{1}{u}h(u))}=c$
Is the aforementioned approach right or wrong?
Any help is appreciated,
Thanks in advance.

Re: First order ODEs
Q3:
I also get $\displaystyle \arctan(r^2)=\arctan(s)+C \Leftrightarrow \arctan(r^2)+\arctan(s)=C$
Now, I would do the following:
$\displaystyle \tan[ \arctan(r^2)+\arctan(s)]=\tan(C)$
Expand the LHS by using $\displaystyle \tan(a+b)=\frac{\tan(a)+\tan(b)}{1\tan(a)\cdot \tan(b)}$ and using the fact that $\displaystyle \tan[\arctan(x)]=x$

Re: First order ODEs
Thanks Siron, I managed to understand Q3. Do you have any idea about the other questions especially Q14?

Re: First order ODEs
For 14 you may try
$\displaystyle x+y=p$
$\displaystyle xy=q$.

Re: First order ODEs
Thanks, I will try the substitution you mentioned. Any other suggestions?