 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2015/discrete-mathematics-continuous-mathematics-and-probability-resit/5/b/solution

Since

x > - 1

, divide by

x + 1

y^{'} - \frac{3}{x + 1} y = (x + 1)^{4} .

μ (x) = e^{- 3 l o g (x + 1)} = (x + 1)^{- 3}

. Multiplication by this factor gives

\frac{d}{d x} [(x + 1)^{- 3} y] = x + 1.

(x + 1)^{- 3} y = \frac{1}{2} (x + 1)^{2} + C

, so

y (x) = \frac{1}{2} (x + 1)^{5} + C (x + 1)^{3}, C \in R, x > - 1.

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