First slide

Introduction to Differential equations

Question

$If \frac{d y}{d x} + y \sec x = \tan x then (\sqrt{2} + 1) y (\frac{π}{4}) - y (o) =$

Easy

A

$\sqrt{2} - \frac{π}{4}$
B

$\sqrt{2} + \frac{π}{4}$
C

$\sqrt{2} - \frac{π}{2}$
D

$\sqrt{2} + \frac{π}{2}$

Solution

$\begin{array}{l} I. F = \exp \int \sec x d x = \sec x + \tan x \\ y (\sec x + \tan x) = \int \tan x (\sec x + \tan x) \\ = \sec x + \tan x - x + c \\ x = \frac{π}{4} \to (\sqrt{2} + 1) y (\frac{π}{4}) = \sqrt{2} + 1 - \frac{π}{4} + c \\ x = 0 \to y (0) = 1 + c \\ \Rightarrow (\sqrt{2} + 1) y (\frac{π}{4}) - y (0) = \sqrt{2} - \frac{π}{4} \end{array}$

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