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?
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 $$ \frac{3 x(t) \ln x(t)}{t^2 - 3t} = \frac{27^{1 - 2/t} \ln 9}{t^2} \text{.} $$