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

Question:

If $y=y(x)$ is the solution of the differential equation $\frac{d y}{d x}+(\tan x) y=\sin x, 0 \leq x \leq \frac{\pi}{3}$, with $\mathrm{y}(0)=0$, then $\mathrm{y}\left(\frac{\pi}{4}\right)$ equal to :

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

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

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

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

Correct Option: , 2

Solution:

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

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

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

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

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

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

$y \sec x=\ell \mathrm{n}|\sec x|$

$y=\cos x \cdot \ln |\sec x|$

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

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