Solution to the following Non linear Ode We have the following ODE:
$x'(t) = \frac{3x(t)ln(x(t))}{t^2-3t}$
$x(t_0) = x_0 ∈ R$
My professor claims that the solution is, for $t_0 = 6$ and $x_0 = 3$, $x(t) = 9e^{\frac{− 6 ln 3}{t}}$. Its derivative is $9e^{\frac{− 6 ln 3}{t}}\frac{6 ln 3}{t^2}$. But if we plug our solution $x(t)$ into $x'(t) = \frac{3x(t)ln(x(t))}{t^2-3t}$, we get a different function. How is this possible?
 A: There is at least an error in the work for the Laplace transform of the integral in $t$.  Since both of these integrals are fairly simple, let's just do them.
\begin{align*}
\int \frac{\mathrm{d}x}{3 x \ln x} &= \frac{1}{3} \int \frac{\mathrm{d}u}{u}  &  &\left[u = \ln x, \mathrm{d}u = \frac{\mathrm{d}x}{x} \right]  \\
&= \frac{1}{3} \ln |u| + C_1  \\
&= \frac{1}{3} \ln |\ln x| + C_1
\end{align*}
and
\begin{align*}
\int \frac{\mathrm{d}t}{t^2 - 3t} &= \int \frac{1/3}{t-3} - \frac{1/3}{t} \,\mathrm{d}t  \\
&= \frac{1}{3} \ln |t-3| - \frac{1}{3}\ln |t| + C_2    \\
&= \frac{1}{3} \ln \left| \frac{t-3}{t} \right| + C_2  
\text{.}
\end{align*}
Equating these two results, combining constants of integration, and multiplying through by $3$, we arrive at
\begin{align*}
\ln |\ln x(t)| &= \ln \left| \frac{t-3}{t} \right| + C  \\
&= \ln \left| \frac{t-3}{t} \right| + \ln \mathrm{e}^C  \\
&= \ln \left( \left| \frac{t-3}{t} \right|  \mathrm{e}^C \right)  \text{.}
\end{align*}
The real valued logarithm is injective (also called one-to-one), so
$$  |\ln x(t)| = \left| \frac{t-3}{t} \right| \mathrm{e}^C  \text{.}  $$
As long as $t > 3$, the fraction on the right is positive, so we may remove the absolute value bars.    Since we are working on the connected part of the domain containing the initial condition $(t_0,x_0) = (6,3)$, we must have $t > 3$.
$$  |\ln x(t)| = \frac{t-3}{t} \mathrm{e}^C  \text{.}  $$
This right-hand side is always positive, so, since $x(t)$ is continuous, the sign of $\ln x(t)$ is always positive or always negative.  At the given initial condition, $\ln(x_0) = \ln 3 > 0$, so $\ln x(t) > 0$ for the particular solution we are seeking.  Thus, we may remove the absolute value bars from the left-hand side.
$$  \ln x(t) = \frac{t-3}{t} \mathrm{e}^C  \text{.}  $$
Applying initial conditions immediately, we have 
$$  \ln 3 = \frac{6-3}{6} \mathrm{e}^C  $$
so
$$  2 \ln 3 = \mathrm{e}^C  $$
and we obtain
$$ x(t) = \exp \left( \frac{t-3}{t} 2 \ln 3 \right)  \text{.}  $$
Now we check.  Using that $t > 3$ during simplification, we find
$$  x'(t) = \frac{27^{1 - 2/t} \ln 9}{t^2}  $$
and 
$$  \frac{3 x(t) \ln x(t)}{t^2 - 3t} = \frac{27^{1 - 2/t} \ln 9}{t^2}  \text{.}  $$
