Step 1: Resolve the fraction into partial fractions. We want to resolve $\frac{6}{(1+2x)(3-x)}$ into partial fractions. Set up the partial fraction decomposition: $$ \frac{6}{(1+2x)(3-x)} = \frac{A}{1+2x} + \frac{B}{3-x} $$ Multiply both sides by $(1+2x)(3-x)$: $$ 6 = A(3-x) + B(1+2x) $$ To find $A$, set $1+2x=0$, which means $x = -\frac{1}{2}$: $$ 6 = A\left(3 - \left(-\frac{1}{2}\right)\right) + B(0) $$ $$ 6 = A\left(3 + \frac{1}{2}\right) $$ $$ 6 = A\left(\frac{7}{2}\right) $$ $$ A = \frac{12}{7} $$ To find $B$, set $3-x=0$, which means $x = 3$: $$ 6 = A(0) + B(1+2(3)) $$ $$ 6 = B(1+6) $$ $$ 6 = 7B $$ $$ B = \frac{6}{7} $$ So, the partial fraction decomposition is: $$ \frac{6}{(1+2x)(3-x)} = \frac{12}{7(1+2x)} + \frac{6}{7(3-x)} $$ a) The partial fraction decomposition is $\boxed{\frac{12}{7(1+2x)} + \frac{6}{7(3-x)}}$. Step 2: Solve the differential equation. The given differential equation is $(1+2x)(3-x)\frac{dy}{dx} = 6(y-2)$. Rearrange the equation to separate variables: $$ \frac{dy}{y-2} = \frac{6}{(1+2x)(3-x)} dx $$ Using the result from part (a), substitute the partial fraction decomposition: $$ \frac{dy}{y-2} = \left(\frac{12}{7(1+2x)} + \frac{6}{7(3-x)}\right) dx $$ Integrate both sides: $$ \int \frac{1}{y-2} dy = \int \left(\frac{12}{7(1+2x)} + \frac{6}{7(3-x)}\right) dx $$ $$ \ln|y-2| = \frac{12}{7} \int \frac{1}{1+2x} dx + \frac{6}{7} \int \frac{1}{3-x} dx $$ 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: $$ \ln|y-2| = \frac{12}{7} \left(\frac{1}{2}\ln|1+2x|\right) + \frac{6}{7} (-\ln|3-x|) + C $$ $$ \ln|y-2| = \frac{6}{7}\ln|1+2x| - \frac{6}{7}\ln|3-x| + C $$ Combine the logarithmic terms on the right side: $$ \ln|y-2| = \frac{6}{7}\ln\left|\frac{1+2x}{3-x}\right| + C $$ Step 3: Apply the initial condition to find $C$. Given that $y=3$ when $x=0$: $$ \ln|3-2| = \frac{6}{7}\ln\left|\frac{1+2(0)}{3-0}\right| + C $$ $$ \ln(1) = \frac{6}{7}\ln\left|\frac{1}{3}\right| + C $$ $$ 0 = \frac{6}{7}\ln\left(\frac{1}{3}\right) + C $$ Since $\ln\left(\frac{1}{3}\right) = \ln(3^{-1}) = -\ln(3)$: $$ 0 = \frac{6}{7}(-\ln(3)) + C $$ $$ C = \frac{6}{7}\ln(3) $$ Step 4: Write the particular solution. Substitute the value of $C$ back into the general solution: $$ \ln|y-2| = \frac{6}{7}\ln\left|\frac{1+2x}{3-x}\right| + \frac{6}{7}\ln(3) $$ $$ \ln|y-2| = \frac{6}{7}\left(\ln\left|\frac{1+2x}{3-x}\right| + \ln(3)\right) $$ Using the logarithm property $\ln a + \ln b = \ln(ab)$: $$ \ln|y-2| = \frac{6}{7}\ln\left|3 \cdot \frac{1+2x}{3-x}\right| $$ Using the logarithm property $k\ln a = \ln(a^k)$: $$ \ln|y-2| = \ln\left|\left(\frac{3(1+2x)}{3-x}\right)^{6/7}\right| $$ Exponentiate both sides to remove the logarithm: $$ |y-2| = \left|\left(\frac{3(1+2x)}{3-x}\right)^{6/7}\right| $$ 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. $$ y-2 = \left(\frac{3(1+2x)}{3-x}\right)^{6/7} $$ $$ y = 2 + \left(\frac{3(1+2x)}{3-x}\right)^{6/7} $$ b) The solution to the differential equation is $\boxed{y = 2 + \left(\frac{3(1+2x)}{3-x