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

જો  \(y=y(x)\) એ વિકલ સમીકરણ \(\frac{d y}{d x}+y \tan x=x \sec x, \quad 0 \leq x \leq \frac{\pi}{3}\), \(y (0)=1\) નો ઉકેલ હોય  તો \(y \left(\frac{\pi}{6}\right)\) ની કિમંત મેળવો.

  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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