Question

A function $y=f\left(x\right)$ satisfying the differential equation $\frac{dy}{dx}.\sin x-y\cos x+\frac{{\sin }^{2}x}{x^2}=0$ such that $y\to 0$ as $x\to \infty$ then the correct statement which is correct is

• A. $\underset{x\to 0}{lim}itf\left(x\right)=1$
• B. $\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx$ is less than $\frac{\pi }{2}$
• C. $\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx$ is greater than unity
• D. $f\left(x\right)$ is an odd function

Answer 1

Answer: (B)

$\underset{0}{\overset{\pi /2}{\int }}f\left(x\right)dx$ is less than $\frac{\pi }{2}$

$\frac{dy}{dx}-y\cot x=\frac{-\sin x}{x^2}$ {dividing by $\sin x$)

Integral function (IF)$=e^{\int -\cot xdx}=e^{-\log \sin x}=\frac{1}{\sin x}$

Multiplying by $\frac{1}{\sin x}$, we get

$\frac{1}{\sin x}\times \frac{dy}{dx}-y\frac{\cos x}{{\sin }^{2}x}=\frac{-1}{x^2}$

$d\left(\frac{1}{\sin x}y\right)=\frac{-1}{x^2}dx$

now, integrate, we get...

$\frac{y}{\sin x}=\frac{1}{x}+C$

$y=\frac{\sin x}{x}+c\sin x$

Now applying condition, ${lim}_{x\to \infty }y=0,$ we get $C=0$

So, $y=f\left(x\right)=\frac{\sin x}{x}$

Since, $y$ is even function,

${\int }_0^{\frac{\pi }{2}}y=\frac{1}{2}{\int }_0^{\pi }\frac{\sin x}{x}=I$ (Let)

On calculating, we found that once,

$\frac{\sin x}{x}<1,x\in R,I<\frac{\pi }{2}$