This question was on my quiz for Calc AB, everyone was confused on how to solve it and I did some random substitution but was still getting nowhere. My teacher said it's possible to do it but not possible with the knowledge we have currently and she's going to throw out the problem. The problem was as following:
$$\frac{\mathrm{d}y}{\mathrm{d}x}=y-3x$$
Well, just set $y=t+3x+3$ to get that $$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d}t}{\mathrm{d}x}+3=y-3x=t+3$$
From $y=3x+3$.
So we have $$\frac{\mathrm{d}t}{\mathrm{d}x}=t$$ So $t=ce^{x}$ for some constant $c$, as shown here. Since $y=t+3x+3$, we get that $y=ce^{x}+3x+3$ for some constant $c$.
Although I'm not in the US, this certainly isn't among course material introduced here.
It is doable, but indeed not with the material from Calc AB. First you need to look at the general solution of your DFQ, which is solving $y'=y$ which is standard to be $y=Ae^x$. The particular solution is found by substituting $y=mx+b$ into the DFQ to obtain $m=mx+b-3x$ equivalent to $0=(m-3)x+b-m.$ From here you can find $m=3$ and $b=3$. The solution then follows $y=Ae^x+3x+3$
Multiply by $e^{-x}$: $$y'e^{-x}-ye^{-x}=-3xe^{-x}$$ Integrating we obtain: $$ye^{-x}=3(x+1)e^{-x}+C$$ $$y=3x+3+Ce^{-x}$$
However, in this exceptional case it is possible to solve it using the method as shown by @S.C.B.
– projectilemotion Feb 24 '17 at 16:45