Explanation:
(A) : Given,
Differential equation \(\frac{d y}{d x}+2 y \tan x=\sin x\),
The given differential equation is of the form
\(\frac{d y}{d x}+p y=Q \text { on comparing, we get }\)
\(p=2 y \tan x \text { and } Q=\sin x\)
Now, We fill find the integrating factor-
\(\text { IF }=e^{\int \operatorname{ddx}}=\mathrm{e}^{\int 2 \tan x d x}\)
\(=\mathrm{e}^{2} \int \tan x d x=e^{[\log \sec x]=e^{\log \left(\sec ^{2} x\right)}}=\sec ^{2} x\)
\(\therefore \quad\) Solution to the differential equation-
\(y(\mathrm{IF})=\int(\mathrm{Q} \times \mathrm{IF}) \mathrm{dx}+\mathrm{c}\)
\(\mathrm{y} \sec ^{2} \mathrm{x}=\int \sin x \sec ^{2} x d x+c\)
\(y \sec ^{2} x=\int \frac{\sin x}{\cos ^{2} x} d x+c=\int \frac{\sin x}{\cos x} \times \frac{1}{\cos x} d x+c\)
\(y \sec ^{2} x=\int \tan x \sec x d x+c\)
\(y \sec ^{2} x=\sec x+c\)
\(y=\frac{1}{\sec x}+\frac{c}{\sec ^{2} x}=\cos x+\left(\cos ^{2} x\right) c\)
When, \(x=\frac{\pi}{3}, y=0\)
\(y=\cos \left(\frac{\pi}{3}\right)+\cos ^{2}\left(\frac{\pi}{3}\right) c\)
\(=\frac{1}{2}+\frac{1}{4}(c)=0=\frac{2+c}{4}=0 \text { or } 2+c=0\)
\(\therefore \quad \mathrm{c}=-2\)
\(\frac{\mathrm{y}}{2}=-\cos ^{2} \mathrm{x}+\frac{1}{2} \cos \mathrm{x}-2 \cos ^{2} \mathrm{x}\)
\(\frac{\mathrm{y}}{2}=-\left[\left(\cos ^{2} \mathrm{x}-\frac{1}{2} \cos \mathrm{x}\right)=-\left[\left(\cos \mathrm{x}-\frac{1}{4}\right)^{2}+\left(\frac{1}{4}\right)^{2}\right]\right.\)
\(\left.\left(\cos \mathrm{x}-\frac{1}{4}\right)^{2}+\left(\frac{1}{4}\right)^{2}\right]=-\left(\cos \mathrm{x}-\frac{1}{4}\right)^{2}+\frac{1}{16}\)
\(\left.\cos \mathrm{x}-\frac{1}{4}\right)^{2} \quad \text { will always be possible and therefore- }\)
\(\left(\cos \mathrm{x}-\frac{1}{4}\right)^{2} \leq 0\)
\(\text { Hence, will be max. value of }\left(\frac{1}{4}\right)^{2}=\frac{1}{16}\)
\(\text { So, The max. value of } \mathrm{y}=\frac{1}{16} \times 2=\frac{1}{8}\)
