Let y = y(x) be the solution of the different | vedclass

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

Question

(B) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$.This can be written as $\frac{d}{dx}(y \sin x) = 4x$.Integrating both si

Vedclass · Accepted Answer

$-\frac{8}{9} \pi^2$

Vedclass · Answer

(B) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$.This can be written as $\frac{d}{dx}(y \sin x) = 4x$.Integrating both sides with respect to $x$,we get $y \sin x = \int 4x \, dx = 2x^2 + C$.Given that $y(\frac{\pi}{2}) = 0$,we substitute $x = \frac{\pi}{2}$ and $y = 0$ into the equation:$0 \cdot \sin(\frac{\pi}{2}) = 2(\frac{\pi}{2})^2 + C \Rightarrow 0 = 2(\frac{\pi^2}{4}) + C \Rightarrow C = -\frac{\pi^2}{2}$.Thus,the solution is $y \sin x = 2x^2 - \frac{\pi^2}{2}$.Now,we find $y(\frac{\pi}{6})$ by substituting $x = \frac{\pi}{6}$:$y \sin(\frac{\pi}{6}) = 2(\frac{\pi}{6})^2 - \frac{\pi^2}{2}$.$y(\frac{1}{2}) = 2(\frac{\pi^2}{36}) - \frac{\pi^2}{2} = \frac{\pi^2}{18} - \frac{\pi^2}{2} = \frac{\pi^2 - 9\pi^2}{18} = -\frac{8\pi^2}{18} = -\frac{4\pi^2}{9}$.Therefore,$y = 2 	imes (-\frac{4\pi^2}{9}) = -\frac{8\pi^2}{9}$.

Let $y = y(x)$ be the solution of the differential equation $\sin x \frac{dy}{dx} + y \cos x = 4x$,where $x \in (0, \pi)$. If $y(\frac{\pi}{2}) = 0$,then find the value of $y(\frac{\pi}{6})$.

Similar Questions

$A$ value of $n$ such that $\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)^n=1$ is

$A$ crystal of intrinsic silicon at room temperature has a carrier concentration of $1.6 \times 10^{16} / m^3$. If the donor concentration level is $4.8 \times 10^{20} / m^3$,then the concentration of holes in the semiconductor is

Which of the following has the lowest boiling point?

What are $X$ and $Y$ in the following reaction?
$CF_2Cl_2 \xrightarrow{UV} X + Y$

$\int \frac{x^{49} \tan ^{-1}\left(x^{50}\right)}{1+x^{100}} d x = k\left(\tan ^{-1}\left(x^{50}\right)\right)^2 + c$,then $k$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module