The solution of the differential equation cos | vedclass

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

Question

(D) Given the differential equation: $\cos y \log(\sec x + 	an x) dx = \cos x \log(\sec y + 	an y) dy$. Separating the variables,we get: $\frac{\log

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Given the differential equation: $\cos y \log(\sec x + 	an x) dx = \cos x \log(\sec y + 	an y) dy$. Separating the variables,we get: $\frac{\log(\sec x + 	an x)}{\cos x} dx = \frac{\log(\sec y + 	an y)}{\cos y} dy$. Since $\frac{1}{\cos x} = \sec x$,the equation becomes: $\sec x \log(\sec x + 	an x) dx = \sec y \log(\sec y + 	an y) dy$. Integrating both sides: $\int \sec x \log(\sec x + 	an x) dx = \int \sec y \log(\sec y + 	an y) dy$. Let $u = \log(\sec x + 	an x)$,then $du = \frac{1}{\sec x + 	an x} (\sec x 	an x + \sec^2 x) dx = \frac{\sec x(	an x + \sec x)}{\sec x + 	an x} dx = \sec x dx$. Similarly,for the right side,let $v = \log(\sec y + 	an y)$,then $dv = \sec y dy$. The integral becomes: $\int u du = \int v dv$. This yields $\frac{u^2}{2} = \frac{v^2}{2} + C_1$,or $u^2 - v^2 = C$. Substituting back,we get: $[\log(\sec x + 	an x)]^2 - [\log(\sec y + 	an y)]^2 = C$. Comparing this with the given options,none of them match. Therefore,the correct option is $(d)$.

The solution of the differential equation $\cos y \log(\sec x + \tan x) dx = \cos x \log(\sec y + \tan y) dy$ is

Similar Questions

$a, b, c, d$ are real numbers. The general solution of $\frac{dy}{dx} = \frac{ax+b}{cy+d}$ represents a family of straight lines,when

The general solution of the differential equation $\frac{d y}{d x}+\frac{\sin (2 x+y)}{\cos x}+2=0$ is:

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

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

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module