If y = y ( x ) is the solution curve of the differential equation d y d x + y tan x = x sec x , 0 x 3 , y ( 0 ) = 1 , then y ( 6 ) is equal to [2023]

Question

If $y = y (x)$ is the solution curve of the differential equation $\frac{d y}{d x} + y \tan x = x \sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$ , then $y (\frac{π}{6})$ is equal to [2023]

Smart Achievers · Accepted Answer

(1)

$\frac{d y}{d x} + y \tan x = x \sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$

$I.F. = e^{\int \tan x d x} = e^{\log_{e} \sec x} = \sec x$ $\Rightarrow y \cdot \sec x = \int x \sec^{2} x d x$

$\Rightarrow y \sec x = x \tan x - \int \tan x d x$

$\Rightarrow y \cdot \sec x = x \tan x - \log_{e} (\sec x) + c$

$Now y (0) = 1 \Rightarrow 1 = c$

$Now, at x = \frac{π}{6}, we have$

$\frac{2 y}{\sqrt{3}} = \frac{π}{6} \cdot \frac{1}{\sqrt{3}} - \log_{e} \frac{2}{\sqrt{3}} + 1;$ $y = \frac{π}{12} - \frac{\sqrt{3}}{2} \log_{e} \frac{2}{\sqrt{3}} + \frac{\sqrt{3}}{2}$

$= \frac{π}{12} - \frac{\sqrt{3}}{2} [\log_{e} \frac{2}{\sqrt{3}} - \log_{e} e]$ $= \frac{π}{12} - \frac{\sqrt{3}}{2} \log_{e} (\frac{2}{e \sqrt{3}})$

Solution

Q.
If $y = y (x)$ is the solution curve of the differential equation $\frac{d y}{d x} + y \tan x = x \sec x, 0 \leq x \leq \frac{π}{3}, y (0) = 1$ , then $y (\frac{π}{6})$ is equal to [2023]

1 $\frac{π}{12} - \frac{\sqrt{3}}{2} \log_{e} (\frac{2}{e \sqrt{3}})$

2 $\frac{π}{12} + \frac{\sqrt{3}}{2} \log_{e} (\frac{2}{e \sqrt{3}})$

3 $\frac{π}{12} + \frac{\sqrt{3}}{2} \log_{e} (\frac{2 \sqrt{3}}{e})$

4 $\frac{π}{12} - \frac{\sqrt{3}}{2} \log_{e} (\frac{2 \sqrt{3}}{e})$

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Solution

Q. If y=y(x) is the solution curve of the differential equation dydx+ytanx=xsecx, 0≤x≤π3, y(0)=1, then y(π6) is equal to [2023]

1 π12-32loge(2e3)

2 π12+32loge(2e3)

3 π12+32loge(23e)

4 π12-32loge(23e)