S2.2 Q1

[Yifei Hu]

The problem asks for general solution of the equation $U_t+yU_x-xU_y=0; U(0,x,y)=f(x,y)$
I proceed as usual:
$$\frac{dt}{1}=\frac{dx}{y}=\frac{-dy}{x}=\frac{du}{0}$$
Integrate and this gives: $x^2+y^2=C$, $t-\int \frac{1}{\sqrt{c-x^2}}dx=D$
Hence, I conclude that $U=\phi(C,D)=\phi(x^2+y^2,t-\int \frac{1}{\sqrt{c-x^2}}dx)$. However, this involves an integral that I can not calculated by hand, can anyone give me a hint on how to do this integral?
Also, I see that in the solution we can also solve this system with a nice trigonometry form $U=f(xcos(t)-ysin(t),xsin(t)+ycos(t))$but the solution does not specify how to reach that, can anyone shed lights on how the solution is reached?
Thanks in advance.

[Victor Ivrii]

• If you do not know this integral you need to refresh Calcuus I. one of basic integrals. Or have a table of basic integrals handy.
• Since $x^2+y^2=c^2$ is a circle, you can substitute $x=c\cos(s)$ and $y=c\sin(s)$ and then observe that $s=D-s$. It gives you the answer, less nicely looking than the one you wrote.
• Expressing $x, y$ through $t,c,d$ you can express $C=c\cos(d)$ and $D=c\sin(d)$ through $x,y,t$ which would give you that nice answer.
  

Write \cos , \sin , \log .... to produce proper (upright) expressions with proper spacing