If (2 + sin x) frac{dy}{dx} + (y + 1) cos x = | vedclass

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

Question

(B) Given the differential equation: $(2 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.This can be rewritten as: $\frac{d}{dx} [(2 + \sin x)(y + 1)] =

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(B) Given the differential equation: $(2 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$.This can be rewritten as: $\frac{d}{dx} [(2 + \sin x)(y + 1)] = 0$.Integrating both sides with respect to $x$,we get: $(2 + \sin x)(y + 1) = C$,where $C$ is a constant.Using the initial condition $y(0) = 1$,we substitute $x = 0$ and $y = 1$:$(2 + \sin 0)(1 + 1) = C \Rightarrow (2 + 0)(2) = C \Rightarrow C = 4$.Thus,the equation becomes: $(2 + \sin x)(y + 1) = 4$.Solving for $y$: $y + 1 = \frac{4}{2 + \sin x} \Rightarrow y = \frac{4}{2 + \sin x} - 1$.Now,we find $y(\frac{\pi}{2})$:$y(\frac{\pi}{2}) = \frac{4}{2 + \sin(\frac{\pi}{2})} - 1 = \frac{4}{2 + 1} - 1 = \frac{4}{3} - 1 = \frac{1}{3}$.

If $(2 + \sin x) \frac{dy}{dx} + (y + 1) \cos x = 0$ and $y(0) = 1$,then $y(\frac{\pi}{2}) = \dots$

Similar Questions

The general solution of the differential equation $x y(y+2) dy + (y^3-1) dx = 0$ is

Find the particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ given that $y = 1$ when $x = 0$.

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

The solution of the differential equation $y \ dx - x \ dy = xy \ dx$ is . . . . . .

The general solution of the differential equation $y y^{\prime} = x \left[ \frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi^{\prime}\left(\frac{y^2}{x^2}\right)} \right]$,where $\phi$ is an arbitrary function,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module