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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077당신이 사용하는 게 아니라요?교과서 변경
7장, 문제 7.PE.127

In Exercises 125–128 solve the differential equation.
127. yy' = sec(y²)sec²(x)

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Rewrite the given differential equation \(yy' = \sec(y^{2}) \sec^{2}(x)\) by expressing \(y'\) as \(\frac{dy}{dx}\). This gives \(y \frac{dy}{dx} = \sec(y^{2}) \sec^{2}(x)\).
Separate the variables so that all terms involving \(y\) are on one side and all terms involving \(x\) are on the other side. This means rewriting the equation as \(y \, dy = \sec(y^{2}) \sec^{2}(x) \, dx\) and then rearranging to isolate \(y\) and \(dy\) on one side and \(x\) and \(dx\) on the other.
Look carefully at the terms to see if you can write the equation in the form \(f(y) \, dy = g(x) \, dx\). Since \(\sec(y^{2})\) depends on \(y\), try to express the left side as a function of \(y\) only and the right side as a function of \(x\) only.
Integrate both sides separately: \(\int f(y) \, dy = \int g(x) \, dx\). This will involve integrating \(y\) and \(\sec(y^{2})\) with respect to \(y\), and \(\sec^{2}(x)\) with respect to \(x\).
After integration, include the constant of integration \(C\) and write the implicit solution relating \(y\) and \(x\). If possible, solve for \(y\) explicitly, but often the solution may remain implicit.

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