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I tried an alternative technique to see if it would work, and somewhere I went wrong. The point of the technique is that a slightly simpler version of the problem should be "easier" to solve than the original version.

We start with ##y'(x) = (ax + b)y(x)## then make the substitution ##u = ax + b \implies \frac{u-b}{a} = x##. This is the tricky part now I think I am having trouble with. Is the left-hand side ##y'( \frac{u-b}{a} )## or is the left-hand side ##\frac{1}{a} y'( \frac{u-b}{a})## by the chain rule? I'm not entirely sure any more.

In either case, the next step is to relabel ##y( \frac{u-b}{a})## as ##z(u)##, so it's important that we know what is happening with the original left-hand side because we need to find a relationship for ##\frac{dz}{du}## by differentiating both sides of ##y( \frac{u-b}{a}) = z(u)##.

So the expression, after those couple substitutions, we should end up with an expression like ##z'(u) = \frac{u}{a}z(u)##, which may be incorrect, but then we could explicitly solve for ##z## and then back-substitute to obtain our original function ##y(x)## once we solve for ##z## in terms of ##u## explicitly.

In principle this is slightly easier than the original problem because there is no ##+b## term in the coefficient of ##z##.