Solve the equation $tx'=x(\ln x - \ln t)$
First note the domain $x>0\land t>0$. Next, we write this differential equation into standard form:
$$\frac{dx}{dt}=\frac{x}{t}\cdot\ln\left(\frac{x}{t}\right),~~~~~~~~~t>0\land x>0\land \color{red}{x\neq t}\tag{1}$$
Note for this implicit condition $$\color{red}{x\neq t}$$
otherwise, (1) gives
$$1=1\cdot\ln 1=0$$
Now, let's go back to (1). This is a homogeneous differential equation, which has the standard form:
$$\frac{dx}{dt}=F\left(\frac{x}{t}\right)$$
so we take the standard substitution $x=u\cdot t$ for solving homogeneous differential equations. The idea for this standard substitution is to convert the original differential equation into a separable equation. Note that $u=u(t)$ is function of $t$, by product rule, we get
$$\frac{dx}{dt}=t\frac{du}{dt}+u$$
Now, eq.(1) becomes:
$$t\frac{du}{dt}+u=u\ln u,~~~~~~~t>0\land u>0\land\color{red}{u\neq 1}\tag{2}$$
Note that the red-color restriction in (1) is equivalent to
$$\color{red}{x\neq t}\Longleftrightarrow\color{red}{u\neq 1}$$
Also, note for a trivial solution from (2):
$$\boxed{\color{blue}{u=e}}\tag{3}$$
We can see now this differential equation (2) becomes separable, namely:
$$\frac{du}{u\ln u-u}=\frac{dt}{t}$$
Under the restrictions $t>0\land u>0\land\color{red}{u\neq 1}\land \color{blue}{u\neq e}$, we can integrate both sides,
$$\ln t +C_1=\int\frac{1}{u\ln u-u}du=\int\frac{1}{\ln (u)-1}\cdot\frac{1}u~du$$
Note, in (1) and (2) we require $t>0$, hence we don't need the absolute value for $\ln t$ in the LHS. For RHS, let $\theta=\ln u,~~~\theta\in\mathbb{R}$
$$\ln t +C_1=\int\frac{1}{\theta-1}~d\theta=\ln|\theta-1|+C_2$$
Simplify and we get:
$$|\theta-1|=e^{\ln t}\cdot e^{C_1-C_2}=e^{C_1-C_2}\cdot t$$
Let $C=e^{C_1-C_2}>0$, hence:
$$\boxed{|\ln (u)-1|=C\cdot t,~~~~~t>0\land u>0\land\color{red}{u\neq 1}\land \color{blue}{u\neq e}}\tag{4}$$
So far, we are done for the integration steps. Next, we need to determine the constanst $C$ and the take off the absolute value, i.e. choose either $+$ or $-$ branch. This depends on your initial conditions. By the Existence and Uniqueness theorem, the solution is uniquely determined on the interval where your initial condition belongs to, and before the solution (curve) hits the nearest singularity. In our case, these singularities are those boundary points, such as $t=0, u=0, u=1, u=e$. You can find more details on this existence and uniqueness with my answers on other similar examples here and here.
Finally, we convert (3) and (4) into variable $x$ and $t$
Trivial solution:
$$\boxed{x=et,~~~~~t>0}\tag{5}$$
None trivial solution ($C>0$):
$$\boxed{~\left|\ln \left(\frac{x}t\right)-1\right|=C\cdot t,~~~~~t>0\land \frac{x}t>0\land\color{red}{\frac{x}t \neq 1}\land \color{blue}{\frac{x}t\neq e}~~}\tag{6}$$
$$=====================\text{Remarks}=====================$$
Note in (4) and (6), the entire domain for $u$ and $\frac{x}t$ are splited into three regions:
$$u\in (0,1)\cup(1,e)\cup(e,\infty)\Longleftrightarrow \frac{x}t\in (0,1)\cup(1,e)\cup(e,\infty)$$
By existence and uniqueness theorem, the solution is uniquely determined on one interval of them, which depends on the initial conditions. We choose the implicit form without taking off the absolute value, because this problem didn't specify the initial conditions. As long as the initial conditions are specified, we can immediately convert the implicit form into explicit form.
For example, if the initial condition is $t_0=1, x_0=2$, then we have
$$\frac{x_0}{t_0}=\frac{2}1=2\in (1,e)$$
So we know $\displaystyle \frac{x}t\in (1,e)$, hence
$$0<\ln\left(\frac{x}t\right)<1$$
Now we can take off the absolute value without any ambiguity for signs, hence (6) becomes
$$1-\ln \left(\frac{x}t\right)=C\cdot t,~~~~~t>0\land \frac{x}t\in (1,e)$$
Plug in initial condition $t_0=1, x_0=2$, we get
$$C=1-\ln2>0$$
hence
$$x=t\cdot e^{1-(1-\ln2)\cdot t}\Longrightarrow x=t\cdot 2^t\cdot e^{1-t}$$
