Word Count : 471

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We are asked to solve the differential equation [ y'' + xy = 0 ] using the transformation [ y = x^{1/2} u, \quad 2x^{3/2} = 3z \implies z = \frac{2}{3} x^{3/2}. ] Start with the transformation (y = x^{1/2} u). Then [ y' = \frac{1}{2} x^{-1/2} u + x^{1/2} u', \quad y'' = -\frac{1}{4} x^{-3/2} u + x^{-1/2} u' + x^{1/2} u''. ] Substitute (y, y'') into the equation (y'' + xy = 0): [ \left(-\frac{1}{4} x^{-3/2} u + x^{-1/2} u' + x^{1/2} u'' \right) + x \cdot x^{1/2} u = 0 ] [ -\frac{1}{4} x^{-3/2} u + x^{-1/2} u' + x^{1/2} u'' + x^{3/2} u = 0. ] Divide through by (x^{1/2}) (valid for (x ________ __________ ______ ____ ___ _____ _________ _____ _____ _________.
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