Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 3 / c

Using the causal Green's function,

y_{p} (x) = \int_{0}^{x} G (x, ξ) ξ d ξ = x^{r} \int_{0}^{x} ξ^{2 - r} lo g (x / ξ) d ξ .

Put

ξ = x s

. Since

α > - 5

is equivalent to

r < 3

, the improper integral converges and gives

y_{p} (x) = x^{3} \int_{0}^{1} s^{2 - r} (- lo g s) d s = \frac{x ^{3}}{( 3 - r ) ^{2}} = \frac{4 x ^{3}}{( α + 5 ) ^{2}} .

The elementary identity

\int_{0}^{1} s^{p - 1} (- lo g s) d s = 1/ p^{2}

follows by differentiating

\int_{0}^{1} s^{p - 1} d s = 1/ p

for

p > 0

. Direct substitution checks

L (x^{3}) = (α + 5)^{2} x /4

, and the displayed solution and its first derivative vanish at zero.

To give all solutions allowed by the printed boundary conditions, retain the singular-endpoint qualification from part (b):

y (x) = ⎩ ⎨ ⎧ \frac{4 x ^{3}}{( α + 5 ) ^{2}} + x^{(1 - α) /2} (C_{1} + C_{2} lo g x), \frac{4 x ^{3}}{( α + 5 ) ^{2}}, - 5 < α < - 1, α \geq - 1.

Thus the causal integral gives a particular member of a two-parameter family in the first range and the unique solution in the second.

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