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JEE Mains · Maths · STD 12 - 9. differential equations

यदि अवकल समीकरण \(\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y} \tan \mathrm{x}=\mathrm{x} \sec \mathrm{x}\), \(0 \leq \mathrm{x} \leq \frac{\pi}{3}, \mathrm{y}(0)=1\) का हल वक्र \(\mathrm{y}=\mathrm{y}(\mathrm{x})\) है, तो \(\mathrm{y}\left(\frac{\pi}{6}\right)\) का मान है:

  1. A \(\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right)\)
  2. B \(\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{ e }\left(\frac{2 \sqrt{3}}{ e }\right)\)
  3. C \(\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _{ e }\left(\frac{2 \sqrt{3}}{ e }\right)\)
  4. D \(\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _{ e }\left(\frac{2}{ e \sqrt{3}}\right)\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right)\)

Step-by-step Solution

Detailed explanation

Here I.F. \(=\sec x\) Then solution of D.E : \(y(\sec x)=x \tan x-\ln (\sec x)+c\) \(\text { Given } y(0)=1 \Rightarrow c=1\) \(\therefore \quad y(\sec x)=x \tan x-\ln (\sec x)+1\)…
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