If cos x frac{dy}{dx} - y sin x = 6x,where 0 | vedclass

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

Question

(C) Given the differential equation: $\cos x \frac{dy}{dx} - y \sin x = 6x$. This can be written as the derivative of a product: $\frac{d}{dx}(y \cos

Vedclass · Accepted Answer

$\frac{-\pi^2}{2 \sqrt{3}}$

Vedclass · Answer

(C) Given the differential equation: $\cos x \frac{dy}{dx} - y \sin x = 6x$. This can be written as the derivative of a product: $\frac{d}{dx}(y \cos x) = 6x$. Integrating both sides with respect to $x$: $y \cos x = \int 6x \, dx = 3x^2 + C$. Using the initial condition $y(\frac{\pi}{3}) = 0$: $0 \cdot \cos(\frac{\pi}{3}) = 3(\frac{\pi}{3})^2 + C \Rightarrow 0 = \frac{\pi^2}{3} + C \Rightarrow C = \frac{-\pi^2}{3}$. Thus,the general solution is $y \cos x = 3x^2 - \frac{\pi^2}{3}$. Now,to find $y(\frac{\pi}{6})$,substitute $x = \frac{\pi}{6}$: $y(\frac{\pi}{6}) \cos(\frac{\pi}{6}) = 3(\frac{\pi}{6})^2 - \frac{\pi^2}{3}$. $y(\frac{\pi}{6}) \cdot \frac{\sqrt{3}}{2} = 3(\frac{\pi^2}{36}) - \frac{\pi^2}{3} = \frac{\pi^2}{12} - \frac{4\pi^2}{12} = \frac{-3\pi^2}{12} = \frac{-\pi^2}{4}$. Therefore,$y(\frac{\pi}{6}) = \frac{-\pi^2}{4} \cdot \frac{2}{\sqrt{3}} = \frac{-\pi^2}{2 \sqrt{3}}$.

If $\cos x \frac{dy}{dx} - y \sin x = 6x$,where $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{3}) = 0$,then find $y(\frac{\pi}{6})$.

Similar Questions

Let $y=y(x)$ be the solution curve of the differential equation $\sec y \frac{dy}{dx} + 2x \sin y = x^3 \cos y$,with the initial condition $y(1) = 0$. Then the value of $y(\sqrt{3})$ is equal to:

The solution of the differential equation $ydx - (x + 2y^2)dy = 0$ is $x = f(y)$. If $f(-1) = 1$,then $f(1)$ is equal to

If the function $y = f(x)$ satisfies the differential equation $(x^3 + 1)dy = x(1 - 3xy)dx$ and $f(0) = 0$,then $\mathop {\lim }\limits_{x \to 0} \frac{x^2}{f(x)}$ is equal to

For the differential equation $y^2 dx + \left( x - \frac{1}{y} \right) dy = 0$ with the initial condition $y(1) = 1$,find $x$.

The general solution of the differential equation $(y^2+x+1) dy = (y+1) dx$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module