# If y=y(x) is the solution of the differential

Question:

If $y=y(x)$ is the solution of the differential equation $\frac{\mathrm{dy}}{\mathrm{dx}}+(\tan \mathrm{x}) \mathrm{y}=\sin \mathrm{x}, 0 \leq \mathrm{x} \leq \frac{\pi}{3}$, with

$y(0)=0$, then $y\left(\frac{\pi}{4}\right)$ equal to :

1. $\frac{1}{4} \log _{\mathrm{e}} 2$

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

3. $\log _{\mathrm{e}} 2$

4. $\frac{1}{2} \log _{e} 2$

Correct Option: , 2

Solution:

$\frac{\mathrm{dy}}{\mathrm{dx}}+(\tan \mathrm{x}) \mathrm{y}=\sin \mathrm{x} ; 0 \leq \mathrm{x} \leq \frac{\pi}{3}$

I.F. $=\mathrm{e}^{\int \tan x d x}=\mathrm{e}^{\ln \sec x}=\sec x$

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

$y \sec x=\ell \operatorname{n}|\sec x|+C$

$\mathrm{x}=0, \mathrm{y}=0 \quad \Rightarrow \quad \therefore \mathrm{c}=0$

$y \sec x=\ell n|\sec x|$

$y=\cos x \cdot \ell n|\sec x|$

$\left.\mathrm{y}\right|_{\mathrm{x}=\frac{\pi}{4}}=\left(\frac{1}{\sqrt{2}}\right) \cdot \ell \mathrm{n} \sqrt{2}$

$\left.\mathrm{y}\right|_{\mathrm{x}=\frac{\pi}{4}}=\frac{1}{2 \sqrt{2}} \log _{\mathrm{e}} 2$