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  • If is the solution curve of the differential equation  <img alt="\frac{d y}{d x}+y \tan x=x \sec x, 0 \l

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\left(\frac{\pi}{6}\right) is equal to

Option: 1

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


Option: 2

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


Option: 3

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


Option: 4

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


Answers (1)

best_answer

Given D.E. is linear D.E.
$$ \begin{aligned} & \text { I.F. }=e^{\int \tan x d x} \\ & =e^{\ell \operatorname{nsec} x}=\operatorname{secx} \end{aligned}
Solution is -
$$ \begin{aligned} & y \sec x=\int x \sec ^2 x d x \\ & =\mathrm{x} \tan \mathrm{x}-\int \tan x d x \\ & \Rightarrow \quad y \sec x=x \tan x-\ell \text { n sect }+c \\ & \text { Put } y(0)=1 \\ & 1=0-0+\mathrm{c} \Rightarrow \mathrm{c}=1 \\ & \mathrm{Y}(\mathrm{x})=\frac{x \tan x}{\sec x}-\frac{\ell \mathrm{nsec} x}{\sec x}+\frac{1}{\sec x} \\ & y\left(\frac{\pi}{6}\right)=\frac{\left(\frac{\pi}{6}\right)\left(\frac{1}{\sqrt{3}}\right)}{\left(\frac{2}{\sqrt{3}}\right)}-\frac{\ln \left(\frac{2}{\sqrt{3}}\right)}{\left(\frac{2}{\sqrt{3}}\right)}+\frac{\sqrt{3}}{2} \\ & =\frac{\pi}{12}-\frac{\sqrt{3}}{2} \ell n\left(\frac{2}{\sqrt{3}}\right)+\frac{\sqrt{3}}{2} \ell n e \\ & =\frac{\pi}{12}-\frac{\sqrt{3}}{2} \ell n\left(\frac{2}{e \sqrt{3}}\right) \\ & \end{aligned}

Posted by

Gautam harsolia

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