The general solution of the differential equa | vedclass

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

Question

(C) Given the differential equation: ${e^y}\frac{{dy}}{{dx}} + ({e^y} + 1)\cot x = 0$Rearranging the terms to separate the variables:${e^y}\frac{{dy}}

Vedclass · Accepted Answer

$({e^y} + 1)\sin x = K$

Vedclass · Answer

(C) Given the differential equation: ${e^y}\frac{{dy}}{{dx}} + ({e^y} + 1)\cot x = 0$Rearranging the terms to separate the variables:${e^y}\frac{{dy}}{{dx}} = -({e^y} + 1)\cot x$$\frac{{{e^y}}}{{{e^y} + 1}}dy = -\cot x dx$Integrating both sides:$\int \frac{{{e^y}}}{{{e^y} + 1}}dy = -\int \cot x dx$Let $u = {e^y} + 1$,then $du = {e^y} dy$. The integral becomes:$\int \frac{1}{u} du = -\int \cot x dx$$\ln|{e^y} + 1| = -\ln|\sin x| + C$$\ln|{e^y} + 1| + \ln|\sin x| = C$Using the property $\ln A + \ln B = \ln(AB)$:$\ln|({e^y} + 1)\sin x| = C$Taking the exponential of both sides,we get:$({e^y} + 1)\sin x = K$ (where $K = e^C$ is a constant).

The general solution of the differential equation ${e^y}\frac{{dy}}{{dx}} + ({e^y} + 1)\cot x = 0$ is

Similar Questions

The solution of the differential equation $2x \frac{dy}{dx} - y = 3$ represents a family of

The general solution of the differential equation $\cos x \sin y \, dx + \sin x \cos y \, dy = 0$ is

The general solution of the differential equation $\frac{dy}{dx} = 1 + x + y + xy$ is

The solution of the differential equation $\cos x \cos y \frac{dy}{dx} = - \sin x \sin y$ is

Find the general solution of the differential equation: $x^{5} \frac{dy}{dx} = -y^{5}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module