Word Count : 382

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To solve the differential equation \( y'' + xy = 0 \) using the transformation \( y = x^{1/2}u \), we first need to express \( y'' \) and \( xy \) in terms of \( u \).

Let's start with the transformation:

\[ y = x^{1/2}u \]

Differentiating \( y \) with respect to \( x \), we get:

\[ y' = \frac{1}{2}x^{-1/2}u + x^{1/2}u' \]

\[ y'' = -\frac{1}{4}x^{-3/2}u + \frac{1}{2}x^{-1/2}u' + x^{1/2}u'' \]

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