Stuck solving this differential equation

$$u^2+u+7=\left(u+\dfrac 12\right)^2+\dfrac {27}4$$ Then: $$I= \int \dfrac {du}{(u+\dfrac 12)^2+\dfrac {27}{4}}$$ $$I= \int \dfrac {dv}{v^2+\dfrac {27}{4}}$$ $$I= \dfrac {2}{3 \sqrt 3 }\int \dfrac {dw}{w^2+1}$$ Where $w = \dfrac {2}{3 \sqrt 3} v$. Use the $\arctan$ function. $$\int \dfrac {dw}{w^2+1}=\arctan w +C$$

Well, we can rewrite your equation:

$$\text{y}'\left(x\right)-\text{y}\left(x\right)\left(6x+1\right)=9x^2+3x+4\tag1$$

Now, let:

$$\mu\left(x\right):=\exp\left\{-\int\left(6x+1\right)\space\text{d}x\right\}=\exp\left(-x\left(3x+1\right)\right)\tag2$$

Multiply both sides by $\mu\left(x\right)$, substitute:

$$\frac{\text{d}}{\text{d}x}\left(\exp\left(-x\left(3x+1\right)\right)\right)=-\left(6x+1\right)\exp\left(-x\left(3x+1\right)\right)\tag3$$

And apply the reverse product rule, to end up with

$$\frac{\text{d}}{\text{d}x}\left(\text{y}\left(x\right)\exp\left(-x\left(3x+1\right)\right)\right)=\left(9x^2+3x+4\right)\exp\left(-x\left(3x+1\right)\right)\tag4$$

Integrate both sides with respect to $x$:

$$\text{y}\left(x\right)\exp\left(-x\left(3x+1\right)\right)=\int\left(9x^2+3x+4\right)\exp\left(-x\left(3x+1\right)\right)\space\text{d}x\tag5$$

So:

$$\text{y}\left(x\right)=\frac{1}{\exp\left(-x\left(3x+1\right)\right)}\int\left(9x^2+3x+4\right)\exp\left(-x\left(3x+1\right)\right)\space\text{d}x\tag6$$

And the integral on the RHS of $(6)$ is defined in terms of the error function.