### Differential Equations Question 10

#### Question 10 - 01 February - Shift 1

If $y=y(x)$ is the solution curve of the differential equation $\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}$,

$y(0)=1$, then $y(\frac{\pi}{6})$ is equal to

(1) $\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e(\frac{2}{e \sqrt{3}})$

(2) $\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _e(\frac{2 \sqrt{3}}{e})$

(3) $\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e(\frac{2 \sqrt{3}}{e})$

(4) $\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _e(\frac{2}{e \sqrt{3}})$

## Show Answer

#### Answer: (1)

#### Formula: Linear differential equations of first order

Неге I.F. $=\sec x$ Then solution of D.E : $ y(\sec x)=x \tan x-\ln (\sec x)+c $

Given $y(0)=1 \Rightarrow c=1$ $ \begin{aligned} & \therefore y(\sec x)=x \tan x-\ln (\sec x)+1 \\ & \text { At } x=\frac{\pi}{6}, y=\frac{\pi}{12}+\frac{\sqrt{3}}{2} \ln \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2} \end{aligned} $