 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2004/paper-2/11/i/solution

For

x = e^{z} > 0

, let

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

. The chain rule gives

y^{'} (x) = \frac{Y ^{'} ( z )}{x}, y^{''} (x) = \frac{Y ^{''} ( z ) - Y ^{'} ( z )}{x ^{2}} .

Multiplying each original Cauchy-Euler equation by

x^{2}

therefore gives

(1) Y^{''} + 3 Y^{'} + 2 Y = 0, α = 3, β = 2,

and

(2) Y^{''} + Y = 0, α = 0, β = 1.

The exponential change of variable has made both equations have constant coefficients.

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