We first write the differential equation $$\frac{\dd y}{\dd t}=p\cdot y+q\cdot t$$ in differential form: $$\dd y-p\cdot y\,\dd t=q\cdot t\,\dd t$$ Multiplying both sides of the equation by $e^{-p\, t}$ gives: $$e^{-p\, t}\,\dd y-p\cdot e^{-p\, t}\cdot y\,\dd t=q\cdot t\cdot e^{-p\, t}\,\dd t$$ With the product rule for differentials we can simplify the left-hand side: $$\dd \left(e^{-p\, t}\cdot y\right)=q\cdot t\cdot e^{-p\, t}\,\dd t$$ So: $$e^{-p\, t}\,y=\int q\cdot t\cdot e^{-p\, t}\,\dd t=q\cdot\int t\cdot e^{-p\, t}\,\dd t$$ The integral on the right-hand side can be computed by means of partial integration: $$\begin{aligned} \int t\cdot e^{-p\, t}\,\dd t &=\int t\cdot\left(-\frac{1}{p}e^{-p\, t}\right)'\,\dd t\\ \\ &=t\cdot\left(-\frac{1}{p}\cdot e^{-p\, t}\right)-\int (t)'\cdot\left(-\frac{1}{p}\cdot e^{-p\, t}\right)\,\dd t\\ \\ &=-\frac{t}{p}\cdot e^{-p\, t}+\frac{1}{p}\cdot \int e^{-p\, t}\,\dd t\\ \\ &=-\frac{t}{p}\cdot e^{-p\, t}-\frac{1}{p^2}\cdot e^{-p\, t}+c\\ \end{aligned}$$ for some constant $c$. So, we can write the solution of the ODE as $$y=c\cdot e^{\,p\, t}-\frac{q}{p}\cdot t-\frac{q}{p^2}$$ Note that the solution is equal to the sum of the specific solution $\displaystyle y=-\frac{q}{r}\cdot t-\frac{q}{p^2}$ and the general solution $y=c\cdot e^{\,p\, t}$ of the homogeneous equation $\displaystyle\frac{\dd y}{\dd t}=p\cdot y.$

In the case of a homogeneous first-order linear ODE $y'=p(t)\cdot y$, the function $\e^{-P(t)}$, where $P(t)$ is an antiderivative of $p(t)$, is an integrating factor: the differential form $$\dd y-p(t)\cdot y\,\dd t=0$$ multiplied by this factor is $$\e^{-P(t)}\,\dd y-\e^{-P(t)}\cdot p(t)\cdot y\,d t = 0$$ which can be rewritten as $$\dd\left(\e^{-P(t)}\cdot y\right) = 0$$ Thus: $\e^{-P(t)}\cdot y = c$, where $c$ is an integration constant. So the general solution is $$y = \e^{P(t)}\cdot C$$