The question in the textbook reads:
Consider the PDE:
$$\sin(t)\frac{\partial u}{\partial x} + x \frac{\partial u}{\partial t} = \frac{x}{t}$$
subject to $u = x$ on $t = 4$.
My characteristics can't be solved in terms of elementary functions and I'm stuck with:
$$\frac{dx}{\sin(t)} = \frac{dt}{x} = \frac{du}{x/t}$$
Edited: PDE corrected.
$$\frac{dx}{\sin(t)} = \frac{dt}{x} = \frac{du}{x/t}\qquad\text{is correct}$$ A first characteristic equation comes from solving $\frac{dx}{\sin(t)} = \frac{dt}{x}$ : $$\frac{x^2}{2}+\cos(t)=c_1$$ A second characteristic equation comes from solving $ \frac{dt}{x} = \frac{du}{x/t}$ : $$u-\ln|t|=c_2$$ The general solution of the PDE from the implicit equation $c_2=F(c_1)$ is : $$u=\ln|t|+F\left(\frac{x^2}{2}+\cos(t)\right)$$ $F$ is an arbitrary function.
Condition : $u(x,4)=x$ $$\ln|4|+F\left(\frac{x^2}{2}+\cos(4)\right)=x$$ Let $X=\frac{x^2}{2}+\cos(4)\quad\implies\quad x=\pm\sqrt{2X-2\cos(4)}$ $$F(X)=-\ln(4)\pm\sqrt{2X-2\cos(4)}$$ Now the function $F(X)$ is determined. We put it into the above general solution where $X=\frac{x^2}{2}+\cos(t)$
$F\left(\frac{x^2}{2}+\cos(t)\right)=-\ln(4)\pm\sqrt{2\left(\frac{x^2}{2}+\cos(t)\right)-2\cos(4)}$ $$u=\ln|t|-\ln(4)\pm\sqrt{2\left(\frac{x^2}{2}+\cos(t)\right)-2\cos(4)}$$ $$u=\ln\left|\frac{t}{4}\right| \pm\sqrt{x^2+2\cos(t)-2\cos(4)}$$ The sign of the square root has to be chosen so that the condition $u(x,4)=x$ be satisfied at $t=4$. That is $+$ if the argument is $>0$ or $-$ if the argument is $<0$.