 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2017/paper-1/13/c/solution

Since

1 + x^{2} > 0

for all real

x

, divide the equation by that factor. Its linear coefficient can be integrated by polynomial division:

P (x) = \frac{3 x ^{3}}{1 + x ^{2}} = 3 x - \frac{3 x}{1 + x ^{2}}, \int P (x) d x = \frac{3}{2} x^{2} - \frac{3}{2} ln (1 + x^{2}) .

M = e^{3 x^{2} /2} (1 + x^{2})^{- 3/2}

. Multiplying the divided equation gives the product rule form

\frac{d}{d x} (M y) = \frac{6 x}{( 1 + x ^{2} ) ^{5/2}} .

u = 1 + x^{2}

gives

M y = - \frac{2}{( 1 + x ^{2} ) ^{3/2}} + C, y = e^{- 3 x^{2} /2} [C (1 + x^{2})^{3/2} - 2] .

y (0) = 1

fixes

C - 2 = 1

. The required solution is therefore

y (x) = e^{- 3 x^{2} /2} [3 (1 + x^{2})^{3/2} - 2], x \in R .

No real singularity was introduced by the division or the integrating factor, so the solution extends over the entire real line.

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