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} $
