 Solution

ID: past-exam-of-the-differential-equations-1-course-of-the-university-of-oxford/2021/2/b/i/solution

With the standard constant starting Picard iterate,

y_{0} (x) = 1

, the next two Picard iterates are

y_{1} (x) = 1 + \int_{1}^{x} (s + \frac{s}{1}) d s = x^{2}

and, for

x > 0

y_{2} (x) = 1 + \int_{1}^{x} (s + \frac{s}{s ^{2}}) d s = 1 + \frac{x ^{2} - 1}{2} + lo g x .

Thus

y_{0} (x) = 1, y_{1} (x) = x^{2}, y_{2} (x) = \frac{x ^{2} + 1}{2} + lo g x .

The restriction

x > 0

for

y_{2}

is essential for the integral from

1

x

: the preceding iterate

y_{1}

vanishes at

0

, where the new integrand

s + 1/ s

is undefined.

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