Problem 5a

$$yu_x-xu_y=x\\

\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x}\\

x^2+y^2 = C\\

u = -y + \phi(x^2+y^2)$$

Problem 5b

$$yu_x-xu_y=x^2\\

\frac{dx}{y} = \frac{dy}{-x} = \frac{du}{x^2}\\

x^2+y^2 = C\\

\left\{\begin{array}{2}

x = r\cos\theta\\

y = r\sin\theta\\

\end{array}\right.\\

du = -xdy = -r^2\cos^2\theta d\theta\\

u = - \frac{r^2}{2} \left(\theta + \frac{1}{2}\sin2\theta\right) + \psi(r)$$

Problem 5c. I think there is a typo in equation (14). The coefficient on $u_x$ should be $y$. Yes indeed. V.I.

$$yu_x+xu_y=x\\

\frac{dx}{y} = \frac{dy}{x} = \frac{du}{x}\\

x^2-y^2 = C\\

u = y + \zeta(x^2-y^2)$$

Problem 5d

$$yu_x+xu_y=x^2\\

\frac{dx}{y} = \frac{dy}{x} = \frac{du}{x^2}\\

x^2-y^2 = C,\qquad -\infty < C < \infty$$

For $0 < C < \infty$, let $C = r^2$.

$$\left\{\begin{array}{2}x = r\sec\theta\\

y = r\tan\theta\\

\end{array}\right.\\

du = xdy = r^2\sec^3\theta d\theta\\

u = \frac{r^2}{2} \left(\sec\theta \tan\theta + \ln|\sec\theta+\tan\theta|\right) + \psi(r)$$

For $-\infty < C < 0$, let $C = -r^2$.

$$\left\{\begin{array}{2}

x = r\tan\theta\\

y = r\sec\theta\\

\end{array}\right.\\

du = xdy = r^2\tan^2\theta\sec\theta d\theta = r^2\left(\sec^3\theta - \sec\theta\right)d\theta\\

u = \frac{r^2}{2} \left(\sec\theta \tan\theta - \ln|\sec\theta+\tan\theta|\right) + \phi(r)$$

At ${x = \pm y}$, $u$ is undefined and discontinuous.