Question

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

• A. \(0\)
• B. \(\sqrt{2}\)
• C. \(1\)
• D. \(-1\)

Hint:

For linear differential equations, identify \(P\), find \(I.F.=e^{\int Pdx}\), then multiply the whole equation by the integrating factor.

Answer 1

The Correct Option is B

Concept: This is a linear differential equation of the form: \[ \frac{dy}{dx}+Py=Q \] Its integrating factor is: \[ I.F.=e^{\int P\,dx} \]

Step 1: Given equation: \[ \frac{dy}{dx}+y\tan x=\sec x \] Here: \[ P=\tan x \] and \[ Q=\sec x \]

Step 2: Find the integrating factor. \[ I.F.=e^{\int \tan x\,dx} \] We know: \[ \int \tan x\,dx=\log \sec x \] So: \[ I.F.=e^{\log \sec x} \] \[ I.F.=\sec x \]

Step 3: Multiply the differential equation by \(\sec x\). \[ \sec x\frac{dy}{dx}+y\sec x\tan x=\sec^2x \] The left side becomes: \[ \frac{d}{dx}(y\sec x)=\sec^2x \]

Step 4: Integrate both sides. \[ y\sec x=\int \sec^2x\,dx \] \[ y\sec x=\tan x+C \]

Step 5: Use \(y(0)=1\). \[ 1\cdot \sec 0=\tan 0+C \] \[ 1\cdot 1=0+C \] \[ C=1 \] Thus: \[ y\sec x=\tan x+1 \]

Step 6: Put \(x=\frac{\pi}{4}\). \[ y\left(\frac{\pi}{4}\right)\sec\frac{\pi}{4} = \tan\frac{\pi}{4}+1 \] We know: \[ \sec\frac{\pi}{4}=\sqrt{2} \] and \[ \tan\frac{\pi}{4}=1 \] So: \[ y\left(\frac{\pi}{4}\right)\sqrt{2}=1+1 \] \[ y\left(\frac{\pi}{4}\right)\sqrt{2}=2 \] \[ y\left(\frac{\pi}{4}\right)=\frac{2}{\sqrt{2}} \] \[ y\left(\frac{\pi}{4}\right)=\sqrt{2} \] Therefore, \[ \boxed{\sqrt{2}} \]