The Indicated function $y_1(x)$ is a solution of the given differential equation. Use reduction of order or formula (5) in Section 4.2, $y_2 = y_1(x) \int \frac{e^{-\int P(x)dx}}{y_1^2(x)} dx$ (5) as instructed, to find a second solution $y_2(x)$. $x^2y'' - 3xy' + 5y = 0$; $y_1 = x^2 \cos(\ln(x))$
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The given differential equation is dx + y(x)dx - 3xy + 5y = 0. To rewrite it in standard form, we can rearrange the terms as follows: dx + y(x)dx + 5y - 3xy = 0 Show moreā¦
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