- Let a family of curves be integral curves of a differential equation $y^{\prime}=f(x, y) .$ Let a second family have the property that at each point $P=(x, y)$ the angle from the curve of the first family through $P$ to the curve of the second family through $P$ is $\alpha .$ Show that the curves of the second family are solutions of the differential equation $$ y^{\prime}=\frac{f(x, y)+\tan \alpha}{1-f(x, y) \tan \alpha} $$
- Use the result of the preceding problem to find the curves that form the angle $\pi / 4$ with
(a) all straight lines through the origin;
(b) all circles $x^{2}+y^{2}=c^{2}$
(Simmons, problem 7.10 and 7.11)
I was trying to work out problem 11. The solutions given are 11.(a) $r=c e^{\theta}$ 11.(b) $r=c e^{-\theta}$
My solution: 11. $y=mx$, $f(x,y)=m$, $\tan \frac{\pi}{4}=1$,thus $$\frac{dy}{dx}=\frac{m+1}{1-m}=C_1$$ $y=r \sin\theta$, $x=r \cos\theta$, $dy= \sin\theta dr$, $dx=-r \sin\theta d\theta$, substituting into our differential equation,
\begin{align} \frac{dy}{dx} &= C_1\\ dy &= C_1 dx\\ \sin\theta dr &= C_1 (-r \sin\theta d\theta)\\ \frac{1}{r} dr &= C_2 d\theta\\ \log r &= C_2 \theta\\ r &= Ce^{\theta} \blacksquare\\ \end{align}
Edit: I realized I made an illegal move: it should be $\log r = C_2 \theta +C_3$, thus $r=Ce^{C_2\theta}$, which is umm...
$x^2+y^2=c^2$, $r=c$, $y=r \sin\theta$, $x=r \cos\theta$ \begin{align} \frac{dy}{dx}&=-\frac{x}{\sqrt{c^2-x^2}}\\ &=-\frac{\cos\theta}{\sin\theta}\\ \end{align}
Substituing into the equation from problem 10,
\begin{align} \frac{dy}{dx}&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}\\ \frac{dy}{dx}&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}\\ dy&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}dx\\ dr&=\frac{-\frac{\cos\theta}{\sin\theta}+1}{1+\frac{\cos\theta}{\sin\theta}}(-r)d\theta\\ \end{align}
How am I supposed to continue? Can I pull off the same trick as before? Thanks in advance.
P.S. I read this post Express the differential equation that solves the below problem in polar form