If y = y(x) is the solution curve of the differential equation dy/dx + ytanx = secx,

Question

If y = y(x) is the solution curve of the differential equation \(\frac{dy}{dx}\) + y tan x = x sec x, \(0 \le x \le \frac \pi 3\), y(0) = 1, then \(\left(\frac {\pi}{6}\right)\) is equal to

(1) \(\frac \pi {12} - \frac{\sqrt 3}2 \log_e\left(\frac 2{e\sqrt 3}\right)\)

(2) \(\frac \pi {12} + \frac{\sqrt 3}2 \log_e\left(\frac {2\sqrt 3}e\right)\)

(3) \(\frac \pi {12} - \frac{\sqrt 3}2 \log_e\left(\frac {2\sqrt 3}e\right)\)

(4) \(\frac \pi {12}+\frac{\sqrt 3}2 \log_e\left(\frac 2{e\sqrt 3}\right)\)

Answer 1

Correct option is (1) \(\frac \pi {12} - \frac{\sqrt 3}2 \log_e\left(\frac 2{e\sqrt 3}\right)\)

Here I.F. = sec x

Then solution of D.E :

y(sec x) = x tan x – ln(sec x) + c

Given y(0) = 1 ⇒ c = 1 

\(\therefore\) y(sec x) = x tan x – ln(sec x) + 1