Find the particular solution of the differential equation e^x tan ydx + (2 -e^x) sec² ydy = 0, given that $y = \frac{π}{4}$ when x = 0

Solution

$e^{x} \tan y dx + (2 - e^{x}) \sec^{2} y dy = 0 e^{x} \tan y dx = (e^{x} - 2) \sec^{2} y dy = 0 e^{x} \tan y dx = (e^{x} - 2) \sec^{2} y dy \int \frac{e^{x} dx}{e^{x} - 2} = \int \frac{\sec^{2} y dy}{\tan y} In | e^{x} - 2 | = In | \tan y | + In C In | e^{x} - 2 | = In (C \tan y) e^{x} - 2 = C \tan y Given; x = 0, y = \frac{π}{4} e^{o} - 2 = C \tan (\frac{π}{4}) e^{o} - 2 = - C \tan (\frac{π}{4}) 1 - 2 = C x 1 \Rightarrow C = - 1 ∴ e^{x} - 2 = - \tan e^{x} - 2 + \tan y = 0$

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Differential Equations

Find the particular solution of the differential equation e^x tan ydx + (2 -e^x) sec² ydy = 0, given that $y = \frac{π}{4}$ when x = 0

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Differential Equations

Find the particular solution of the differential equation ex tan ydx + (2 -ex) sec2 ydy = 0, given that y = π4 when x = 0

Some More Questions From Differential Equations Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the particular solution of the differential equation e^x tan ydx + (2 -e^x) sec² ydy = 0, given that $y = \frac{π}{4}$ when x = 0