 Solution

ID: past-exam-of-the-year-1-of-the-computer-science-course-of-the-university-of-oxford/2014/discrete-mathematics-continuous-mathematics-and-probability/4/d/solution

Write

Y (t) = y (e^{t})

, so

t = lo g x

. The chain rule gives

y^{'} (x) = \frac{Y ^{'} ( t )}{x}, y^{''} (x) = \frac{Y ^{''} ( t ) - Y ^{'} ( t )}{x ^{2}} .

Substitution into the Cauchy-Euler equation therefore gives

Y^{''} + Y^{'} - 2 Y = 0

(r - 1) (r + 2) = 0

, so

Y = C_{1} e^{t} + C_{2} e^{- 2 t}

. Returning to

x

gives

y (x) = C_{1} x + C_{2} x^{- 2}, x > 0.

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