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Exercise 6.3 · Q43

Q.Find the particular solution: 3extan⁡y⋅dx+(1+ex)sec⁡2y⋅dy=03e^x\tan y\cdot dx+(1+e^x)\sec^2y\cdot dy=0, when x=0, y=πx=0,\ y=\pi.

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3extan⁡y dx+(1+ex)sec⁡2y dy=03e^x\tan y\,dx+(1+e^x)\sec^2y\,dy=0 separates as sec⁡2ytan⁡ydy=−3ex1+exdx\dfrac{\sec^2y}{\tan y}dy=-\dfrac{3e^x}{1+e^x}dx. Integrating: log⁡(tan⁡y)=−3log⁡(1+ex)+c1\log(\tan y)=-3\log(1+e^x)+c_1, i.e. tan⁡y (1+ex)3=c\tan y\,(1+e^x)^3=c. At x=0,y=πx=0,y=\pi: tan⁡π=0\tan\pi=0, so the left side is 0⋅(1+1)3=00\cdot(1+1)^3=0, forcing c=0c=0. Since (1+ex)3(1+e^x)^3 is never zero, this forces tan⁡y=0\tan y=0 throughout — i.e. the initial point lies exactly on the singular branch y=πy=\pi (a constant f …

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