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

Question:

If $y=y(x)$ is the solution of the differential equation, $\frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{y} \tan \mathrm{x}=\sin \mathrm{x}, \mathrm{y}\left(\frac{\pi}{3}\right)=0$, then the maximum value of the function $\mathrm{y}(\mathrm{x})$ over $\mathbb{R}$ is equal to:

1. (1) 8

2. (2) $\frac{1}{2}$

3. (3) $-\frac{15}{4}$

4. (4) $\frac{1}{8}$

Correct Option: , 4

Solution:

$\frac{d y}{d x}+2 y \tan x=\sin x$

I.F. $=\mathrm{e}^{\int 2 \tan x d x}=\mathrm{e}^{2 \operatorname{tn} \sec x}$

I.F. $=\sec ^{2} x$

y. $\left(\sec ^{2} x\right)=\int \sin x \cdot \sec ^{2} x d x$

y. $\left(\sec ^{2} x\right)=\int \sec x \tan x d x$

y. $\left(\sec ^{2} x\right)=\sec x+C$

$\mathrm{x}=\frac{\pi}{3} ; \mathrm{y}=0$

$\Rightarrow \quad C=-2$

$\Rightarrow \quad y=\frac{\sec x-2}{\sec ^{2} x}=\cos x-2 \cos ^{2} x$

$\mathrm{y}=\mathrm{t}-2 \mathrm{t}^{2} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dt}}=1-4 \mathrm{t}=0 \Rightarrow \mathrm{t}=\frac{1}{4}$

$\therefore \quad \max =\frac{1}{4}-\frac{1}{8}=\frac{2-1}{8}=\frac{1}{8}$