Find the general solution of the differential | vedclass

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

Question

(A) The given differential equation is: $\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}$.Using trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ a

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The given differential equation is: $\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}$.Using trigonometric identities $1-\cos x = 2 \sin^2 \frac{x}{2}$ and $1+\cos x = 2 \cos^2 \frac{x}{2}$,we get:$\frac{dy}{dx} = \frac{2 \sin^2 \frac{x}{2}}{2 \cos^2 \frac{x}{2}} = 	an^2 \frac{x}{2}$.Since $	an^2 	heta = \sec^2 	heta - 1$,we have:$\frac{dy}{dx} = \sec^2 \frac{x}{2} - 1$.Separating the variables,we get:$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$.Thus,the general solution is $y = 2 	an \frac{x}{2} - x + C$.

Find the general solution of the differential equation: $\frac{dy}{dx} = \frac{1-\cos x}{1+\cos x}$

Similar Questions

The solution of the differential equation $\frac{d\theta}{dt} = -k(\theta - \theta_0)$,where $k$ is a constant,is . . . . . .

Let $x=x(t)$ and $y=y(t)$ be solutions of the differential equations $\frac{dx}{dt}+ax=0$ and $\frac{dy}{dt}+by=0$ respectively,where $a, b \in R$. Given that $x(0)=2$,$y(0)=1$,and $3y(1)=2x(1)$,the value of $t$ for which $x(t)=y(t)$ is:

Find the equation of a curve passing through the point $(-2, 3),$ given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2x}{y^2}$.

If the solution of the differential equation $(2x+3y-2)dx+(4x+6y-7)dy=0$ with $y(0)=3$ is $\alpha x+\beta y+3 \log_e|2x+3y-\gamma|=6$,then $\alpha+2\beta+3\gamma$ is equal to

The solution of the differential equation $\frac{dy}{dx} = \frac{1 - 2y - 4x}{1 + y + 2x}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module