Question:medium

The differential equation is $\frac{dy}{dx} + y\tan x = \sec x$ and $y(0) = 1$. Then the value of $y(\frac{\pi}{4}) =$

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For linear equations $y' + Py = Q$, the integrating factor is always $e^{\int P dx}$. Here $P = \tan x$, making it easy!
  • 0
  • $\sqrt{2}$
  • 1
  • -1
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The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
This is a first-order linear differential equation of the form \( \frac{dy}{dx} + Py = Q \). We solve it using an Integrating Factor (I.F.).
Step 2: Key Formula or Approach:
1. \( I.F. = e^{\int P dx} \).
2. Solution: \( y(I.F.) = \int Q(I.F.) dx + c \).
Step 3: Detailed Explanation:
Here, \( P = \tan x \) and \( Q = \sec x \). \[ I.F. = e^{\int \tan x dx} = e^{\ln|\sec x|} = \sec x \] The solution is: \[ y(\sec x) = \int \sec x \cdot \sec x \, dx \] \[ y \sec x = \int \sec^2 x \, dx \implies y \sec x = \tan x + c \] Use \(y(0) = 1\): \[ (1)\sec(0) = \tan(0) + c \implies 1(1) = 0 + c \implies c = 1 \] General solution: \( y \sec x = \tan x + 1 \implies y = \frac{\tan x + 1}{\sec x} = \sin x + \cos x \). At \( x = \pi/4 \): \[ y = \sin(\pi/4) + \cos(\pi/4) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]
Step 4: Final Answer:
The value of \( y(\pi/4) \) is \( \sqrt{2} \).
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