Resolve the fraction into partial fractions.

Step 1: Resolve the fraction into partial fractions. We want to resolve (6)/((1+2x)(3-x)) into partial fractions. Set up the partial fraction decomposition: (6)/((1+2x)(3-x)) = (A)/(1+2x) + (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 = -(1)/(2): 6 = A(3 - (-(1)/(2))) + B(0) 6 = A(3 + (1)/(2)) 6 = A((7)/(2)) A = (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 = (6)/(7) So, the partial fraction decomposition is: (6)/((1+2x)(3-x)) = (12)/(7(1+2x)) + (6)/(7(3-x)) a) The partial fraction decomposition is (12)/(7(1+2x)) + (6)/(7(3-x)). Step 2: Solve the differential equation. The given differential equation is (1+2x)(3-x)(dy)/(dx) = 6(y-2). Rearrange the equation to separate variables: (dy)/(y-2) = (6)/((1+2x)(3-x)) dx Using the result from part (a), substitute the partial fraction decomposition: (dy)/(y-2) = ((12)/(7(1+2x)) + (6)/(7(3-x))) dx Integrate both sides: (1)/(y-2) dy = ((12)/(7(1+2x)) + (6)/(7(3-x))) dx |y-2| = (12)/(7) (1)/(1+2x) dx + (6)/(7) (1)/(3-x) dx For (1)/(1+2x) dx, let u=1+2x, so du=2dx. The integral becomes (1)/(2) (1)/(u) du = (1)/(2)|1+2x|. For (1)/(3-x) dx, let v=3-x, so dv=-dx. The integral becomes - (1)/(v) dv = -|3-x|. Substitute these back into the equation: |y-2| = (12)/(7) ((1)/(2)|1+2x|) + (6)/(7) (-|3-x|) + C |y-2| = (6)/(7)|1+2x| - (6)/(7)|3-x| + C Combine the logarithmic terms on the right side: |y-2| = (6)/(7)|(1+2x)/(3-x)| + C Step 3: Apply the initial condition to find C. Given that y=3 when x=0: |3-2| = (6)/(7)|(1+2(0))/(3-0)| + C (1) = (6)/(7)|(1)/(3)| + C 0 = (6)/(7)((1)/(3)) + C Since ((1)/(3)) = (3^-1) = -(3): 0 = (6)/(7)(-(3)) + C C = (6)/(7)(3) Step 4: Write the particular solution. Substitute the value of C back into the general solution: |y-2| = (6)/(7)|(1+2x)/(3-x)| + (6)/(7)(3) |y-2| = (6)/(7)(|(1+2x)/(3-x)| + (3)) Using the logarithm property a + b = (ab): |y-2| = (6)/(7)|3 · (1+2x)/(3-x)| Using the logarithm property k a = (a^k): |y-2| = |((3(1+2x))/(3-x))^6/7| Exponentiate both sides to remove the logarithm: |y-2| = |((3(1+2x))/(3-x))^6/7| Since y=3 when x=0, y-2=1 > 0. Also, (3(1+2(0)))/(3-0) = (3)/(3) = 1 > 0. Therefore, we can remove the absolute value signs. y-2 = ((3(1+2x))/(3-x))^6/7 y = 2 + ((3(1+2x))/(3-x))^6/7 b) The solution to the differential equation is $y = 2 + (3(1+2x)3-x