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 Equations1st edition chapter 2.2.

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

14.$\displaystyle (\sqrt(x+y)+\sqrt(x-y))dx+(\sqrt(x-y)-\sqrt(x+y))dy=0$

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 $\displaystyle (x^2-3y^2)dx+2xydy=0$

(c)Use this result to solve $\displaystyle (y+\sqrt(x^2+y^2))dx-xdy=0$

In Q3 I found $\displaystyle \arctan(r^2)+\arctan(s)=C$ and unable to reduce this expresion to $\displaystyle r^2+s=c(1-r^2s)$ as given in the answers at the back of the book. In Q14 I got the following expression and unable to proceed further:

$\displaystyle \int(\frac{(\sqrt(1+v)-\sqrt(1-v))dv}{\sqrt(1+v)+\sqrt(1-v)+v\sqrt(1-v)-v\sqrt(1+v)})=\ln|x|+C$ where $\displaystyle v=\frac{y}{x}$.

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.