The solution of the differential equation (1 | vedclass

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

Question

(A) Given the differential equation: $(1 + \cos x)dy = (1 - \cos x)dx$Rearranging the terms to separate the variables,we get:$dy = \frac{1 - \cos x}{1

Vedclass · Accepted Answer

$y = 2	an \frac{x}{2} - x + c$

Vedclass · Answer

(A) Given the differential equation: $(1 + \cos x)dy = (1 - \cos x)dx$Rearranging the terms to separate the variables,we get:$dy = \frac{1 - \cos x}{1 + \cos x} dx$Using the trigonometric identities $1 - \cos x = 2\sin^2 \frac{x}{2}$ and $1 + \cos x = 2\cos^2 \frac{x}{2}$,we have:$dy = \frac{2\sin^2 \frac{x}{2}}{2\cos^2 \frac{x}{2}} dx = 	an^2 \frac{x}{2} dx$Using the identity $	an^2 	heta = \sec^2 	heta - 1$,we get:$dy = (\sec^2 \frac{x}{2} - 1) dx$Integrating both sides:$\int dy = \int (\sec^2 \frac{x}{2} - 1) dx$$y = 2	an \frac{x}{2} - x + c$

The solution of the differential equation $(1 + \cos x)dy = (1 - \cos x)dx$ is

Similar Questions

If the differential equation $\frac{dy}{dx} + \frac{x}{y} = \frac{a}{y}$,where $a$ is a constant,represents a family of circles,then the radius of the circle is ......

The general solution of the equation $({e^y} + 1)\cos x \, dx + {e^y}\sin x \, dy = 0$ is

The solution of the differential equation $x dy + y dx - \sqrt{1 - x^2 y^2} dx = 0$ is

The particular solution of the differential equation $x dy + 2y dx = 0$,when $x = 2$ and $y = 1$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module