First order ODEs

**JohnDoe** · October 2nd 2011, 01:12 AM

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. $2r(s^2+1)dr+(r^4+1)ds=0$
14. $(\sqrt(x+y)+\sqrt(x-y))dx+(\sqrt(x-y)-\sqrt(x+y))dy=0$
23.(a)Prove that if $Mdx+Ndy=0$ is a homogeneous equation, then the change of variables $x=uy$ transforms this equation into a separable equation in the variables u and x.
(b)Use this result to solve $(x^2-3y^2)dx+2xydy=0$
(c)Use this result to solve $(y+\sqrt(x^2+y^2))dx-xdy=0$

In Q3 I found $\arctan(r^2)+\arctan(s)=C$ and unable to reduce this expresion to $r^2+s=c(1-r^2s)$ as given in the answers at the back of the book. In Q14 I got the following expression and unable to proceed further:
$\int(\frac{(\sqrt(1+v)-\sqrt(1-v))dv}{\sqrt(1+v)+\sqrt(1-v)+v\sqrt(1-v)-v\sqrt(1+v)})=\ln|x|+C$ where $v=\frac{y}{x}$ .
In Q23 I started with $\frac{dy}{dx}=h(x/y)$ then letting x=uy leads $\frac{dy}{dx}=\frac{1}{u}-\frac{x}{u^2}\frac{du}{dx}=h(u)$
It follows that:
$\int\frac{dx}{x}-\int\frac{du}{u^2(\frac{1}{u}-h(u))}=c$
Is the aforementioned approach right or wrong?

Any help is appreciated,

Thanks in advance.

**Siron** · October 2nd 2011, 01:43 AM

Q3:
I also get $\arctan(r^2)=-\arctan(s)+C \Leftrightarrow \arctan(r^2)+\arctan(s)=C$
Now, I would do the following:
$\tan[ \arctan(r^2)+\arctan(s)]=\tan(C)$

Expand the LHS by using $\tan(a+b)=\frac{\tan(a)+\tan(b)}{1-\tan(a)\cdot \tan(b)}$ and using the fact that $\tan[\arctan(x)]=x$

**JohnDoe** · October 2nd 2011, 03:21 AM

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

**zzzoak** · October 2nd 2011, 05:30 AM

For 14 you may try

$x+y=p$

$x-y=q$ .

**JohnDoe** · October 2nd 2011, 08:20 AM

Thanks, I will try the substitution you mentioned. Any other suggestions?

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