Question:medium

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

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For linear differential equations, identify \(P\), find \(I.F.=e^{\int Pdx}\), then multiply the whole equation by the integrating factor.
  • \(0\)
  • \(\sqrt{2}\)
  • \(1\)
  • \(-1\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Question:
The given equation \(\frac{dy}{dx} + y \tan x = \sec x\) is a first-order linear differential equation.
The general form is \(\frac{dy}{dx} + P(x)y = Q(x)\).
To solve this, we need to find the Integrating Factor (I.F.) and then the general solution.
Step 2: Key Formula or Approach:
Identify \(P(x) = \tan x\) and \(Q(x) = \sec x\).
Integrating Factor \(\text{I.F.} = e^{\int P(x) dx}\).
The general solution is \(y \cdot (\text{I.F.}) = \int [Q(x) \cdot (\text{I.F.})] dx + c\).
Step 3: Detailed Explanation:

Step A: Calculate Integrating Factor.
\[ \int P(x) dx = \int \tan x dx = \ln|\sec x| \]
\[ \text{I.F.} = e^{\ln|\sec x|} = \sec x \]

Step B: Find General Solution.
\[ y \cdot \sec x = \int (\sec x \cdot \sec x) dx \]
\[ y \sec x = \int \sec^2 x dx \]
\[ y \sec x = \tan x + c \] ... (1)

Step C: Apply Initial Condition \(y(0) = 1\).
\[ 1 \cdot \sec(0) = \tan(0) + c \]
\[ 1 \cdot 1 = 0 + c \implies c = 1 \]

Step D: The particular solution is \(y \sec x = \tan x + 1\).

Rewriting for \(y\): \(y = \frac{\tan x + 1}{\sec x} = \frac{\frac{\sin x}{\cos x} + 1}{\frac{1}{\cos x}} = \sin x + \cos x\).

Step E: Evaluate at \(x = \frac{\pi}{4}\).
\[ y\left(\frac{\pi}{4}\right) = \sin\left(\frac{\pi}{4}\right) + \cos\left(\frac{\pi}{4}\right) \]
\[ y\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

Step 4: Final Answer:
The solution simplifies to \(y = \sin x + \cos x\). At \(\pi/4\), both terms are \(1/\sqrt{2}\), making the sum \(\sqrt{2}\).
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