Resolve the fraction into partial fractions.

The given differential equation is $(1+2x)(3-x)\frac{dy}{dx} = 6(y-2)$.
Rearrange the equation to separate variables:

Using the result from part (a), substitute the partial fraction decomposition:

Integrate both sides:

For $\int \frac{1}{1+2x} dx$, let $u=1+2x$, so $du=2dx$. The integral becomes $\frac{1}{2}\int \frac{1}{u} du = \frac{1}{2}\ln|1+2x|$.
For $\int \frac{1}{3-x} dx$, let $v=3-x$, so $dv=-dx$. The integral becomes $-\int \frac{1}{v} dv = -\ln|3-x|$.
Substitute these back into the equation:

Combine the logarithmic terms on the right side:

Given that $y=3$ when $x=0$:

Since $\ln\left(\frac{1}{3}\right) = \ln(3^{-1}) = -\ln(3)$:

Substitute the value of $C$ back into the general solution:

Using the logarithm property $\ln a + \ln b = \ln(ab)$:

Using the logarithm property $k\ln a = \ln(a^k)$:

Exponentiate both sides to remove the logarithm:

Since $y=3$ when $x=0$, $y-2=1 > 0$. Also, $\frac{3(1+2(0))}{3-0} = \frac{3}{3} = 1 > 0$. Therefore, we can remove the absolute value signs.