If the solution of the differential equation $(1 + x^3)\dfrac{dy}{dx} + 6x^2y = 1 + x^2$ is $y = \dfrac{1}{(1 + x^3)^s}\left[x + \dfrac{x^p}{p} + \dfrac{x^q}{q} + \dfrac{x^r}{r} + c\right]$, then the LCM of $p, q, r$ and $s$ is...

The given differential equation is $(1 + x^3)\dfrac{dy}{dx} + 6x^2y = 1 + x^2$.

Dividing by $(1 + x^3)$, we get a linear differential equation of the form $\dfrac{dy}{dx} + P(x)y = Q(x)$:

$\dfrac{dy}{dx} + \dfrac{6x^2}{1 + x^3}y = \dfrac{1 + x^2}{1 + x^3}$

The integrating factor (IF) is given by:

$\text{IF} = e^{\int P(x) dx} = e^{\int \dfrac{6x^2}{1 + x^3} dx}$

Let $1 + x^3 = t$, then $3x^2 dx = dt$, which gives $6x^2 dx = 2 dt$.

$\int \dfrac{6x^2}{1 + x^3} dx = \int \dfrac{2}{t} dt = 2 \ln t = \ln(t^2) = \ln((1 + x^3)^2)$

Thus, $\text{IF} = e^{\ln((1 + x^3)^2)} = (1 + x^3)^2$.

Multiplying the differential equation by the IF, we get:

$\dfrac{d}{dx} \left[ y (1 + x^3)^2 \right] = \dfrac{1 + x^2}{1 + x^3} \times (1 + x^3)^2$

$\dfrac{d}{dx} \left[ y (1 + x^3)^2 \right] = (1 + x^2)(1 + x^3)$

$\dfrac{d}{dx} \left[ y (1 + x^3)^2 \right] = 1 + x^2 + x^3 + x^5$

Integrating both sides with respect to $x$:

$y (1 + x^3)^2 = \int (1 + x^2 + x^3 + x^5) dx$

$y (1 + x^3)^2 = x + \dfrac{x^3}{3} + \dfrac{x^4}{4} + \dfrac{x^6}{6} + c$

$y = \dfrac{1}{(1 + x^3)^2} \left[ x + \dfrac{x^3}{3} + \dfrac{x^4}{4} + \dfrac{x^6}{6} + c \right]$

Comparing this with the given solution $y = \dfrac{1}{(1 + x^3)^s}\left[x + \dfrac{x^p}{p} + \dfrac{x^q}{q} + \dfrac{x^r}{r} + c\right]$, we find:

$s = 2$

The set of values for $\{p, q, r\}$ is $\{3, 4, 6\}$.

We need to find the LCM of $p, q, r$ and $s$, which is the LCM of $3, 4, 6,$ and $2$.

$\text{LCM}(2, 3, 4, 6) = 12$

Answer: $12$