Express \(\(2)/((1+x)(1+3x)) ) in partial fractions. Hence solve the differential equation \(\(dy)/(dx) = (2y)/((1+x)(1+3x)) ), given that \(y=1 ) when \(x=0 ).

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Asked on April 16, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-04-16T23:47:40.513Z

You're on a roll — It seems there was some confusion with the numbering in our previous exchanges, as the image you sent contains questions 1 through 5. I have already solved Question 1 (parts a and b), Question 3, and Question 4 ii) from this image. Let's continue by solving Question 2 from the image. Note that Question 4 i) is identical to Question 2. --- Question 2: Express (2)/((1+x)(1+3x)) in partial fractions. Hence solve the differential equation (dy)/(dx) = (2y)/((1+x)(1+3x)), given that y=1 when x=0. Part 1: Partial Fraction Decomposition Step 1: Set up the partial fraction form. (2)/((1+x)(1+3x)) = (A)/(1+x) + (B)/(1+3x) Step 2: Multiply by the common denominator to clear fractions. 2 = A(1+3x) + B(1+x) Step 3: Solve for A and B using substitution. To find A, let 1+x=0 x=-1: 2 = A(1+3(-1)) + B(1-1) 2 = A(1-3) + B(0) 2 = -2A A = -1 To find B, let 1+3x=0 x=-(1)/(3): 2 = A(1+3(-(1)/(3))) + B(1-(1)/(3)) 2 = A(1-1) + B((2)/(3)) 2 = B((2)/(3)) B = 2 × (3)/(2) B = 3 Step 4: Write the partial fraction decomposition. (2)/((1+x)(1+3x)) = (-1)/(1+x) + (3)/(1+3x) The partial fraction decomposition is (3)/(1+3x) - (1)/(1+x). Part 2: Solve the Differential Equation Step 1: Separate the variables. Given the differential equation (dy)/(dx) = (2y)/((1+x)(1+3x)). (1)/(y) (dy)/(dx) = (2)/((1+x)(1+3x)) (1)/(y)\,dy = (2)/((1+x)(1+3x))\,dx Step 2: Substitute the partial fractions into the integral. (1)/(y)\,dy = ((3)/(1+3x) - (1)/(1+x))\,dx Step 3: Integrate both sides. |y| = (3)/(1+3x)\,dx - (1)/(1+x)\,dx For (3)/(1+3x)\,dx, let u = 1+3x, du = 3\,dx. (1)/(u)\,du = |u| = |1+3x| For (1)/(1+x)\,dx, let v = 1+x, dv = \,dx. (1)/(v)\,dv = |v| = |1+x| So, |y| = |1+3x| - |1+x| + C Step 4: Combine the logarithmic terms. Using the logarithm property a - b = ((a)/(b)): |y| = |(1+3x)/(1+x)| + C Step 5: Apply the initial condition y=1 when x=0 to find C. |1| = |(1+3(0))/(1+0)| + C 0 = |(1)/(1)| + C 0 = (1) + C 0 = 0 + C C = 0 Step 6: Write the particular solution. |y| = |(1+3x)/(1+x)| Exponentiate both sides to solve for y: e^|y| = e^|(1+3x)/(1+x)| |y| = |(1+3x)/(1+x)| Since y=1 when x=0, y is positive, so we can remove the absolute value signs for y. Assuming 1+3x and 1+x have the same sign around x=0: y = (1+3x)/(1+x) The solution to the differential equation is y = (1+3x)/(1+x). 3 done, 2 left today. You're making progress.

Accepted Answer · 2026-04-16T23:47:40.513Z

You're on a roll — It seems there was some confusion with the numbering in our previous exchanges, as the image you sent contains questions 1 through 5. I have already solved Question 1 (parts a and b), Question 3, and Question 4 ii) from this image. Let's continue by solving Question 2 from the image. Note that Question 4 i) is identical to Question 2. --- Question 2: Express (2)/((1+x)(1+3x)) in partial fractions. Hence solve the differential equation (dy)/(dx) = (2y)/((1+x)(1+3x)), given that y=1 when x=0. Part 1: Partial Fraction Decomposition Step 1: Set up the partial fraction form. (2)/((1+x)(1+3x)) = (A)/(1+x) + (B)/(1+3x) Step 2: Multiply by the common denominator to clear fractions. 2 = A(1+3x) + B(1+x) Step 3: Solve for A and B using substitution. To find A, let 1+x=0 x=-1: 2 = A(1+3(-1)) + B(1-1) 2 = A(1-3) + B(0) 2 = -2A A = -1 To find B, let 1+3x=0 x=-(1)/(3): 2 = A(1+3(-(1)/(3))) + B(1-(1)/(3)) 2 = A(1-1) + B((2)/(3)) 2 = B((2)/(3)) B = 2 × (3)/(2) B = 3 Step 4: Write the partial fraction decomposition. (2)/((1+x)(1+3x)) = (-1)/(1+x) + (3)/(1+3x) The partial fraction decomposition is (3)/(1+3x) - (1)/(1+x). Part 2: Solve the Differential Equation Step 1: Separate the variables. Given the differential equation (dy)/(dx) = (2y)/((1+x)(1+3x)). (1)/(y) (dy)/(dx) = (2)/((1+x)(1+3x)) (1)/(y)\,dy = (2)/((1+x)(1+3x))\,dx Step 2: Substitute the partial fractions into the integral. (1)/(y)\,dy = ((3)/(1+3x) - (1)/(1+x))\,dx Step 3: Integrate both sides. |y| = (3)/(1+3x)\,dx - (1)/(1+x)\,dx For (3)/(1+3x)\,dx, let u = 1+3x, du = 3\,dx. (1)/(u)\,du = |u| = |1+3x| For (1)/(1+x)\,dx, let v = 1+x, dv = \,dx. (1)/(v)\,dv = |v| = |1+x| So, |y| = |1+3x| - |1+x| + C Step 4: Combine the logarithmic terms. Using the logarithm property a - b = ((a)/(b)): |y| = |(1+3x)/(1+x)| + C Step 5: Apply the initial condition y=1 when x=0 to find C. |1| = |(1+3(0))/(1+0)| + C 0 = |(1)/(1)| + C 0 = (1) + C 0 = 0 + C C = 0 Step 6: Write the particular solution. |y| = |(1+3x)/(1+x)| Exponentiate both sides to solve for y: e^|y| = e^|(1+3x)/(1+x)| |y| = |(1+3x)/(1+x)| Since y=1 when x=0, y is positive, so we can remove the absolute value signs for y. Assuming 1+3x and 1+x have the same sign around x=0: y = (1+3x)/(1+x) The solution to the differential equation is y = (1+3x)/(1+x). 3 done, 2 left today. You're making progress.

Express \(\(2)/((1+x)(1+3x)) ) in partial fractions. Hence solve the differential equation \(\(dy)/(dx) = (2y)/((1+x)(1+3x)) ), given that \(y=1 ) when \(x=0 ).

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Express \(\(2)/((1+x)(1+3x)) ) in partial fractions. Hence solve the differential equation \(\(dy)/(dx) = (2y)/((1+x)(1+3x)) ), given that \(y=1 ) when \(x=0 ).

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