JEE Mains · Maths · STD 12 - 9. differential equations
Let \(y=y(x)\) be the solution of the differential equation \(\cos x\left(\log _{ e }(\cos x)\right)^2 dy +\) \((\sin x~-\) \(3 y \sin x\) \(\log _{ e }(\cos x)) d x=\) \(0, x \in\left(0, \frac{\pi}{2}\right)\). If \(y\left(\frac{\pi}{4}\right)=\frac{-1}{\log _{ e } 2},\) then \(y\left(\frac{\pi}{6}\right)\) is equal to :
- A \(\frac{1}{\log _e(3)-\log _e(4)}\)
- B \(\frac{2}{\log _e(3)-\log _e(4)}\)
- C \(\frac{1}{\log _e(4)-\log _e(3)}\)
- D \(-\frac{1}{\log _e(4)}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{\log _e(3)-\log _e(4)}\)
Step-by-step Solution
Detailed explanation
\(\frac{d y}{d x}-\frac{3 \sin x}{\cos x \ln \cos x} y=-\frac{\sin x}{\cos x(\ln \cos x)^2} \) \(\text {I.F. }=e^{-\int \frac{3 \tan x}{\ln \cos x} d x} \) \(\text {Let } \ln \cos x=t \) \(-\tan x d x=d t \) \(e^{3 \int^{\frac{d t}{t}}}=e^{3 \ln t}=t^3=(\ln \cos x)^3 \)…
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