Consider the differential equation:
\[(4+6t^2) y' + 12 t y = \sin(t).\]We first check if
\[\frac{d}{dt}(4+6t^2) \overset{?}{=} 12 t,\]which is the case.
This means the left-hand side is
\[\left( (4+6t^2) y\right)' = (4+6t^2)y' + 12ty\]by the chain rule.
Hence, combining these two observations
\[\left( (4+6t^2)y\right)' = \sin(t).\]We can integrate both sides to get
\[\begin{split} (4+6t^2)y &= \int \sin(t)\,dt = -\cos(t)+C\\ y&= \frac{-\cos(t)+C}{4+6t^2}. \end{split}\]Is the solution to the differential equations.
Consider the differential equation:
\[y' + \frac{4}{t+3} y = t^2+6t+9.\]We normally multiply by $\mu$ to begin, but we will actually begin by first factoring the right-hand side
\[y' + \frac{4}{t+3}y = (t+3)^2.\]We now multiply by $\mu$, the integrating factor:
\[\mu y' + \frac{4\mu}{t+3} y = (t+3)^2\]and we need the left-hand side to be
\[(\mu y)' = \mu y' + \mu' y = \mu y' + \frac{4\mu}{t+3} y.\]So
\[\mu' = \frac{4\mu}{t+3}.\]We solve this as a separable differential equation:
\[\begin{split} \frac{d\mu}{\mu} &= \frac{4}{t+3} \,dt\\ \int \frac{d\mu}{\mu} &= 4 \int \frac{1}{t+3} \,dt\\ \ln \mu &= 4\ln (t+3) = \ln\left( (t+3)^4\right)\\ \mu &= (t+3)^4. \end{split}\]Where here we ignore the constants of integration because it will not be useful for the general solution to the differential equation we want to solve.
We are now at:
\[\left((t+3)^4 y\right)' = (t+3)^4 (t+3)^2 = (t+3)^6.\]To undo the derivative of the left-hand side we integrate both sides of the equation to get \((t+3)^4 y = \int(t+3)^6\,dt.\)
Using $u = t+3$, we get
\((t+3)^4 y = \frac{1}{7}(t+3)^7 + C\) and rearranging:
\[y = \frac{(t+3)^3}{7} + \frac{C}{(t+3)^4}.\]Consider
\[y' + \frac{6t}{t^2 + 6} y = t-4.\]We begin by multiplying everything by $\mu$, to get
\[\mu y' + \frac{6t}{t^2+6}\,\mu y = (t-4)\mu.\]We want the left-hand side to be just $(\mu y)’ = \mu y’+\mu’ y$.
This means we need
\[\begin{split} \mu' &= \frac{6t}{t^2+6} \mu\\ \frac{d\mu}{\mu} &= \frac{6t}{t^2+6}\,dt\\ \int\frac{d\mu}{\mu} &= 3\int \frac{2t}{t^2+6}\,dt\\ \ln \mu&= 3\ln (t^2+6) = \ln\left((t^2+6)^3\right)\\ \mu &= \left(t^2+6\right)^3. \end{split}\]Again, we ignore the constant of integration at this point in the process of solving the ODE.
We therefore have
\[\left( (t^2+6)^3 y\right)' = (t-4)(t^2+6)^3.\]To undo the derivative we need to integrate both sides of the equation
\[(t^2+6)^3 y = \int (t-4)(t^2+6)^3\,dt.\]We expand the cubed term and then multiply by $t-4$ gives the integrand of the right-hand side is
\[t^7-4t^6+18t^5 -72t^4 +108t^3 -432 t^2+216t - 864.\]We can integrate this to get
\[\begin{split} (t^2+6)^3y &= \frac{1}{8}t^8 - \frac{4}{7}t^7 + 3 t^6 - \frac{72}{5}t^5 \\ &\qquad + 27 t^4 - 144 t^3 + 108 t^2 - 864 t + C. \end{split}\]as an implicit solution.
Solving for $y$ gives
\[\begin{split} y &= \frac{1}{(t^2+6)^3}\bigg( \frac{1}{8}t^8 - \frac{4}{7}t^7 + 3 t^6 - \frac{72}{5}t^5 \\ &\quad \quad+ 27 t^4 - 144 t^3 + 108 t^2 - 864 t + C\bigg). \end{split}\]