Vtyjkyuk

I was browsing through some questions on Method of Characteristics and came across a couple solutions involving characteristic curves:
1. Solve Transport Equation using Method of Characteristics
2. Solve the PDE by method of characteristics.

This method involving the "system of characteristic differential equations" is different to the approach I was taught, and it makes a lot more sense.

The method we were told involves writing out a differential for each segment of the PDE, solving it, and applying boundary and/or initial conditions when necessary. It becomes messy and there are terms popping up, I don't really understand it, despite attempting to go through numerous example questions. I decided to study the 2 questions above.

My problem is with the second one, I just cannot understand it.

First family of characteristic curves, from $quad fracdtt^2=fracdxx^2 quadtoquad frac1t-frac1x=c_1$

My workout led to a different solution:

$$dx=fracx^2t^2dt$$
$$x=intfracx^2t^2dt$$

This allows me to divide both sides by $x^2$. Integrating $t^-2$ gives $-t^-1+C_1$. My answer is:

$$frac1x+frac1t=C_1$$

Then comes the second bit:

Second family of characteristic curves, from $quad fracdtt^2=fracdxx^2= fracfrac1tdt+frac1xdxfrac1tt^2+frac1xx^2 = fracfrac1tdt+frac1xdxt+x=fracdu(t+x)u$
$frac1tdt+frac1xdx=fracduu quadtoquad fracutx=c_2$

Which I do not understand at all.

I decided to do some algebra, and it seems that the pile comes from this:
$$fracdxx+fracdtt=0$$
Then dividing both sides by $(x+t)$, combining the two fractions, and subsequently equating it with $fracdu(t+x)u$.

I then tried to solve the equation:

$$fractdx+xdttx(x+t)=fracdu(t+x)u$$

Which gave me some natural logarithmic.
I multiplied both sides by $t+x$, and then integrated both sides:

$$fracdxx+fracdtt=fracduu$$
$$log(x)+log(t)-log(u)=C_2$$
Edit: I think this is wrong. I cannot integrate the equation with regards to 3 different differentials, right?

Raising this mess as the power of $e$ gets rid of the logs.
And, I am stuck. This looks completely different to the answer, and clearly wrong.

The first issue was likely a typo, since the negative sign was not carried through the calculations.

Any help regarding to the question is greatly appreciated. I tried looking online for material regarding to this approach and did not find anything useful. All the materials I can find does not involve "system of characteristic differential equations" but rather goes straight to various differentials. I think the second step has something to do with separation of variables.

Thanks!

First family: “This allows me to divide both sides by $x^2$.” No, you can't. The variables $x$ and $t$ are not independent, so you can't take $x^2$ outside of the $t$ integral.

What you have is
$$dt/t^2= dx/x^2 iff d(-1/t)=d(-1/x) iff d(1/t-1/x)=0 iff 1/t-1/x=c_1.$$

Second family: If $A=B$, then $(tA+xB)/(t+x)=(tA+xA)/(t+x)=A(t+x)/(t+x)=A$ (which also equals $B$). That is,
$$
A=B
quadimpliesquad
A = B = fractA+xBt+x
.
$$
The computation starts out by applying this identity with $A=dt/t^2$ and $B=dx/x^2$.

And then you get
$$
dt/t + dx/x = du/u
iff
d ln|t| + d ln|x| = d ln|u|
iff
d ln|tfracuxt|=0
iff
tfracuxt=c_2
.
$$
(If the logarithm of the absolute value of something is constant, then that something itself is constant too.)