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

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

Question

(D) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$. Dividing by $\sin x$,we get $\frac{dy}{dx} + y \cot x = \frac{4x}{\sin

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The given differential equation is $\sin x \frac{dy}{dx} + y \cos x = 4x$. Dividing by $\sin x$,we get $\frac{dy}{dx} + y \cot x = \frac{4x}{\sin x}$. This is a linear differential equation of the form $\frac{dy}{dx} + Py = Q$,where $P = \cot x$ and $Q = \frac{4x}{\sin x}$. The integrating factor ($I$.$F$.) is $e^{\int P dx} = e^{\int \cot x dx} = e^{\ln|\sin x|} = \sin x$. The general solution is $y \cdot (I.F.) = \int Q \cdot (I.F.) dx + C$. Substituting the values,$y \sin x = \int \frac{4x}{\sin x} \cdot \sin x dx + C = \int 4x dx + C = 2x^2 + C$. Thus,$y = \frac{2x^2 + C}{\sin x}$. Given $y(\frac{\pi}{2}) = 0$,we have $0 = \frac{2(\frac{\pi}{2})^2 + C}{\sin(\frac{\pi}{2})} = \frac{\pi^2}{2} + C$,so $C = -\frac{\pi^2}{2}$. The particular solution is $y = \frac{2x^2 - \frac{\pi^2}{2}}{\sin x}$. Evaluating at $x = \frac{\pi}{6}$,$y(\frac{\pi}{6}) = \frac{2(\frac{\pi}{6})^2 - \frac{\pi^2}{2}}{\sin(\frac{\pi}{6})} = \frac{\frac{2\pi^2}{36} - \frac{\pi^2}{2}}{1/2} = 2 \left( \frac{\pi^2}{18} - \frac{\pi^2}{2} \right) = 2 \left( \frac{\pi^2 - 9\pi^2}{18} \right) = 2 \left( \frac{-8\pi^2}{18} \right) = -\frac{8}{9} \pi^2$.

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

Similar Questions

The solution of the differential equation $ydx - xdy + \log x dx = 0$ is

Let $f: [1, \infty) \rightarrow R$ be a differentiable function such that $f(1) = \frac{1}{3}$ and $3 \int_1^x f(t) dt = x f(x) - \frac{x^3}{3}$ for $x \in [1, \infty)$. Then the value of $f(e)$ is:

The solution of the differential equation $x \log x \frac{dy}{dx} + y = 2 \log x$ is

The integrating factor of the differential equation $x \frac{dy}{dx} + y \log x = x e^x \cdot x^{-1/2} \log x$ for $x > 0$ is:

The solution of the differential equation $(1+x) \frac{dy}{dx} - xy = 1-x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module