The differential equation cot y , dx = x , dy | vedclass

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

Question

(B) Given the differential equation: $\cot y \, dx = x \, dy$Separate the variables:$\frac{dx}{x} = \frac{dy}{\cot y}$Since $\frac{1}{\cot y} = 	an y

Vedclass · Accepted Answer

$x = c \sec y$

Vedclass · Answer

(B) Given the differential equation: $\cot y \, dx = x \, dy$Separate the variables:$\frac{dx}{x} = \frac{dy}{\cot y}$Since $\frac{1}{\cot y} = 	an y$,we have:$\frac{dx}{x} = 	an y \, dy$Integrating both sides:$\int \frac{dx}{x} = \int 	an y \, dy$$\ln |x| = \ln |\sec y| + \ln |c|$Using the property of logarithms $\ln a + \ln b = \ln(ab)$:$\ln |x| = \ln |c \sec y|$Taking the exponential of both sides:$x = c \sec y$

The differential equation $\cot y \, dx = x \, dy$ has a solution of the form

Similar Questions

The solution of the differential equation $y \, dx - x \, dy + x y^2 \, dx = 0$ is:

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

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

Find the general solution of the differential equation: $e^{x} \tan y \, dx + (1 - e^{x}) \sec^{2} y \, dy = 0$.

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module