Seperation of variables different solutions

Question

Let's consider the ODE $$t\cdot x'(t)=x(t)$$.

First approach:

$$\frac{x'(t)}{x(t)}=1/t,\quad\text{for }t\ne 0$$

$$\int\frac{x'(t)}{x(t)}dt= \int \frac1 t dt$$

$$\log(x(t))=\log|t|+C$$

$$\iff x(t)=\underbrace{e^C}_{=:C_1}|t|$$.

Second approach:

$dx(t)/dt=x(t)/t$ for $t\ne 0$ \begin{align} \int \frac 1 x dx = \int \frac 1 t dt&\iff \log|x(t)|=\log|t| + C\\ &\iff |x(t)|=\underbrace{e^C}_{=C_1}|t|\\ &\iff x(t)=C_1t. \end{align}

You see only the second approach is the solution of this ODE. Why do we not obtain the same result in both approaches?

Answer 1

For any real valued function we have that

$$\int\frac{x'(t)}{x(t)}\,\mathrm dt=\ln(|x(t)|)+ C$$

Then if $x$ is real-valued your first solution is wrong.

Answer 2

The antiderivative of $1/x$ is either $\log x$ where $x$ is positive or $\log(-x)$ where $x$ is negative. In either of those two cases, your two solutions are equivalent.

You will need to analyze what happens at zero separately, since both of your methods are invalid (involve dividing by zero) there.

In particular, you will need to specify whether you allow a solution with discontinuous derivative. And, if you do: both of your solutions are incomplete, the complete solution being $$ x(t) = \begin{cases} C_1 t, \quad t > 0,\\ C_2 t, \quad t < 0 \end{cases} $$