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Exercise 9.3 · Q10

Q.extan⁡y dx+(1−ex)sec⁡2y dy=0e^x \tan y \, dx + (1 - e^x) \sec^2 y \, dy = 0

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Separable equation; integrating sec⁡2ytan⁡y dy=−ex1−ex dx\frac{\sec^2 y}{\tan y}\,dy=-\frac{e^{x}}{1-e^{x}}\,dx gives tan⁡y=C(1−ex)\tan y=C(1-e^{x}). No numeric initial value is supplied, so CC remains arbitrary.

1. Choose the method

Each coefficient splits into an xx-only factor and a yy-only factor, so the variables separate. There is no y/xy/x structure here, so it is not homogeneous.

2. Separate

extan⁡y dx=−(1−ex)sec⁡2y dy  ⇒  sec⁡2ytan⁡y dy=−ex1−ex dx.e^{x}\tan y\,dx=-(1-e^{x})\sec^2 y\,dy\;\Rightarrow\;\frac{\sec^2 y}{\tan y}\,dy=-\frac{e^{x}}{1-e^{x}}\,dx.

3. Integrate both sides

  • Left: with t=tan⁡y, dt=sec⁡2y dyt=\tan y,\ dt=\sec^2 y\,dy, ∫sec⁡2ytan⁡y dy=log⁡∣tan⁡y∣\displaystyle\int\frac{\sec^2 y}{\tan y}\,dy=\log|\tan y|.
  • Right: with u=1−ex, du=−ex dxu=1-e^{x},\ du=-e^{x}\,dx, −∫ex1−ex dx=∫duu=log⁡∣1−ex∣\displaystyle-\int\frac{e^{x}}{1-e^{x}}\,dx=\int\frac{du}{u}=\log|1-e^{x}|.

Hence

log⁡∣tan⁡y∣=log⁡∣1−ex∣+c1.\log|\tan y|=\log|1-e^{x}|+c_1.

4. Simplify

Exponentiating,

tan⁡y=C (1−ex),C=±ec1.\tan y=C\,(1-e^{x}),\qquad C=\pm e^{c_1}. …

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