A function y = f(x) satisfying the differenti | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The given differential equation is $\frac{dy}{dx} \sin x - y \cos x = -\frac{\sin^2 x}{x^2}$.Dividing by $\sin^2 x$,we get $\frac{1}{\sin x} \frac

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

(D) The given differential equation is $\frac{dy}{dx} \sin x - y \cos x = -\frac{\sin^2 x}{x^2}$.Dividing by $\sin^2 x$,we get $\frac{1}{\sin x} \frac{dy}{dx} - y \frac{\cos x}{\sin^2 x} = -\frac{1}{x^2}$.This is equivalent to $\frac{d}{dx} \left( \frac{y}{\sin x} ight) = -\frac{1}{x^2}$.Integrating both sides,we get $\frac{y}{\sin x} = \int -\frac{1}{x^2} dx = \frac{1}{x} + C$.So,$y = \sin x \left( \frac{1}{x} + C ight)$.Given $y ightarrow 0$ as $x ightarrow \infty$,we must have $C = 0$,so $f(x) = \frac{\sin x}{x}$.$1$. $\mathop {Lim}\limits_{x 	o 0} f(x) = \mathop {Lim}\limits_{x 	o 0} \frac{\sin x}{x} = 1$.$2$. Since $\sin x  0$,$f(x) = \frac{\sin x}{x} $3$. Using the series expansion $\frac{\sin x}{x} = 1 - \frac{x^2}{6} + \frac{x^4}{120} - \dots$,the integral $\int_0^{\pi/2} (1 - \frac{x^2}{6}) dx = [x - \frac{x^3}{18}]_0^{\pi/2} = \frac{\pi}{2} - \frac{\pi^3}{144} \approx 1.57 - 0.21 = 1.36 > 1$.Thus,all statements are correct.

$A$ function $y = f(x)$ satisfying the differential equation $\frac{dy}{dx} \sin x - y \cos x + \frac{\sin^2 x}{x^2} = 0$ is such that $y \rightarrow 0$ as $x \rightarrow \infty$. Which of the following statements is correct?

Similar Questions

Let $y=y(x)$ be the solution of the differential equation $e^{y} \frac{d y}{d x}-2 e^{y} \sin x+\sin x \cos ^{2} x=0$,with $y(\frac{\pi}{2})=0$. If $y(0)=\log _{e}(\alpha+\beta e^{-2})$,then $4(\alpha+\beta)$ is equal to $....$

If $y=\sin x+A \cos x$ is the general solution of $\frac{dy}{dx}+f(x)y=\sec x$,then an integrating factor of the differential equation is

Let $y=y(x)$ be the solution curve of the differential equation $\frac{dy}{dx}+\frac{1}{x^{2}-1}y=\left(\frac{x-1}{x+1}\right)^{\frac{1}{2}}$,$x>1$ passing through the point $\left(2, \sqrt{\frac{1}{3}}\right)$. Then $\sqrt{7}y(8)$ is equal to.

The solution of $dy = \cos x(2 - y \csc x)dx$ where $y = 2$ when $x = \frac{\pi}{2}$ is

The solution of the differential equation $\cos x \, dy = y(\sin x - y) \, dx$ for $0 < x < \frac{\pi}{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module