Calculate the particular solution of dy/dx - y/(x ln x) = 1/x at (2; 0). Determine the general solution of d2y/dx2 + dy/dx − 6y = 2e⁻³x.

$$\frac{dy}{dx} + \left(-\frac{1}{x \ln x}\right)y = \frac{1}{x}$$

Here, $P(x) = -\frac{1}{x \ln x}$ and $Q(x) = \frac{1}{x}$.

First, calculate $\int P(x) dx$:

$$\int -\frac{1}{x \ln x} dx$$

Let $u = \ln x$, then $du = \frac{1}{x} dx$.

$$\int -\frac{1}{u} du = -\ln|u| = -\ln|\ln x|$$

Now, find the integrating factor:

$$I(x) = e^{-\ln|\ln x|} = e^{\ln((\ln x)^{-1})} = \frac{1}{\ln x}$$

(Assuming $x>1$ for $\ln x$ to be positive and defined, which is true for $x=2$).

The left side becomes $\frac{d}{dx}(y \cdot I(x))$.

$$\frac{d}{dx}\left(y \cdot \frac{1}{\ln x}\right) = \frac{1}{x} \cdot \frac{1}{\ln x}$$ $$\frac{d}{dx}\left(\frac{y}{\ln x}\right) = \frac{1}{x \ln x}$$

Integrate both sides with respect to $x$:

$$\int \frac{d}{dx}\left(\frac{y}{\ln x}\right) dx = \int \frac{1}{x \ln x} dx$$ $$\frac{y}{\ln x} = \ln|\ln x| + C$$

The general solution is $y = \ln x (\ln|\ln x| + C)$.

Substitute $x=2$ and $y=0$ into the general solution:

$$0 = \ln 2 (\ln|\ln 2| + C)$$

Since $\ln 2 \neq 0$, we must have:

$$\ln|\ln 2| + C = 0$$ $$C = -\ln|\ln 2|$$

Substitute the value of $C$ back into the general solution:

$$y = \ln x (\ln|\ln x| - \ln|\ln 2|)$$

Using the logarithm property $\ln a - \ln b = \ln\left(\frac{a}{b}\right)$:

$$y = \ln x \left( \ln \left| \frac{\ln x}{\ln 2} \right| \right)$$

The particular solution is:

$$\boxed{y = \ln x \ln\left(\frac{\ln x}{\ln 2}\right)}$$

The characteristic equation is:

$$r^2 + r - 6 = 0$$

Factor the quadratic equation:

$$(r+3)(r-2) = 0$$

The roots are $r_1 = -3$ and $r_2 = 2$.
The complementary solution is:

$$y_c = C_1 e^{-3x} + C_2 e^{2x}$$

The right-hand side is $g(x) = 2e^{-3x}$. Since $-3$ is a root of the characteristic equation (and it's a simple root), we must multiply the standard guess by $x$.
Let $y_p = Ax e^{-3x}$.
Calculate the first and second derivatives of $y_p$:

$$y_p' = A e^{-3x} + Ax(-3e^{-3x}) = A e^{-3x} - 3Ax e^{-3x} = A(1-3x)e^{-3x}$$ $$y_p'' = A(-3e^{-3x}) + (-3A e^{-3x} - 3Ax(-3e^{-3x})) = -3A e^{-3x} - 3A e^{-3x} + 9Ax e^{-3x} = A(-6+9x)e^{-3x}$$

Substitute $y_p$, $y_p'$, and $y_p''$ into the non-homogeneous differential equation:

$$A(-6+9x)e^{-3x} + A(1-3x)e^{-3x} - 6(Ax e^{-3x}) = 2e^{-3x}$$

Divide by $e^{-3x}$:

$$A(-6+9x) + A(1-3x) - 6Ax = 2$$ $$-6A + 9Ax + A - 3Ax - 6Ax = 2$$

Combine like terms:

$$(-6A + A) + (9Ax - 3Ax - 6Ax) = 2$$ $$-5A + 0x = 2$$ $$-5A = 2$$ $$A = -\frac{2}{5}$$

So, the particular solution is:

$$y_p = -\frac{2}{5} x e^{-3x}$$

$$y = C_1 e^{-3x} + C_2 e^{2x} - \frac{2}{5} x e^{-3x}$$

The general solution is:

$$\boxed{y = C_1 e^{-3x} + C_2 e^{2x} - \frac{2}{5} x e^{-3x}}$$