Question

If $\mathrm{y}=\mathrm{y}\left(\mathrm{x}\right)$ is the solution of the differential equation $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)+\left(\mathrm{tanx}\right)\mathrm{y}=\mathrm{sinx},0\le \mathrm{\pi }\le \frac{\mathrm{\pi }}{3},$ with$\mathrm{y}\left(0\right)=0$, then $y\left(\frac{\mathrm{\pi }}{4}\right)$ equal to:

• A. ${\log }_{e}2$
• B. $\left(\frac{1}{2}\right){\log }_{e}2$
• C. $\left(\frac{1}{2\sqrt{2}}\right){\log }_{e}2$
• D. $\left(\frac{1}{4}\right){\log }_{e}2$

Answer 1

The correct option is C

$\left(\frac{1}{2\sqrt{2}}\right){\log }_{e}2$

Explanation for the correct option:

Finding the value of $y\left(\frac{\mathrm{\pi }}{4}\right)$:

The given differential equation is $\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)+\left(\mathrm{tanx}\right)\mathrm{y}=\mathrm{sinx},0\le \mathrm{\pi }\le \frac{\mathrm{\pi }}{3}$

First, find the integrating factor,

$\begin{array}{rcl}\text{I.F}& =& {\mathrm{e}}^{\int \mathrm{tanxdx}}\\ & =& {\mathrm{e}}^{\ln \mathrm{secx}}\\ & =& \mathrm{secx}\end{array}$

The differential equation is

$$\begin{gathered} \mathrm{y}·\mathrm{I}.\mathrm{F}=\int \mathrm{Q}\left(\mathrm{x}\right)·\mathrm{I}.\mathrm{Fdx} \\ \Rightarrow \mathrm{y}\mathrm{secx}=\int \mathrm{tanx}\mathrm{dx} \\ \Rightarrow \mathrm{ysecx}=\ln \left|\mathrm{secx}\right|+\mathrm{C}...\left(1\right) \end{gathered}$$

By substitute $\mathrm{x}=0,\mathrm{y}=0$

$$\begin{gathered} \left(0\right)=\ln \left|0\right|+\mathrm{C} \\ \Rightarrow C=0 \end{gathered}$$

Substitute the value of $\mathrm{C}$ in equation $\left(1\right)$

$$\begin{gathered} \begin{array}{l}\mathrm{y}\mathrm{secx}=\ln \left|\mathrm{secx}\right|\\ \Rightarrow \mathrm{y}=\frac{1}{\mathrm{secx}}.\ln \left|\mathrm{secx}\right|\end{array} \\ =\mathrm{cosx}.\ln \left|\mathrm{secx}\right| \end{gathered}$$

Put $\mathrm{x}=\frac{\mathrm{\pi }}{4}$

$\begin{array}{l}{\left.\mathrm{y}\right|}_{\mathrm{x}=\frac{\mathrm{\pi }}{4}}=\left(\frac{1}{\sqrt{2}}\right).\ln \sqrt{2}\mathbf{[}\mathbf{\because }\mathbf{}\mathbf{cos}\left(\frac{\pi }{4}\right)\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{2}}}\mathbf{]}\\ {\left.\mathrm{y}\right|}_{\mathrm{x}=\frac{\mathrm{\pi }}{4}}=\frac{1}{2\sqrt{2}}{\log }_{\mathrm{e}}2\left[\because {\mathrm{loga}}^b=\mathrm{bloga}\right]\end{array}$

Hence, the correct option is C.