The solution of the differential equation sec | vedclass

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

Question

(B) Given the differential equation: $\sec^2 x 	an y \, dx + \sec^2 y 	an x \, dy = 0$Rearranging the terms to separate the variables:$\sec^2 x 	an

Vedclass · Accepted Answer

$	an x 	an y = \sqrt{3}$

Vedclass · Answer

(B) Given the differential equation: $\sec^2 x 	an y \, dx + \sec^2 y 	an x \, dy = 0$Rearranging the terms to separate the variables:$\sec^2 x 	an y \, dx = -\sec^2 y 	an x \, dy$Dividing both sides by $	an x 	an y$:$\frac{\sec^2 x}{	an x} \, dx = -\frac{\sec^2 y}{	an y} \, dy$Integrating both sides:$\int \frac{\sec^2 x}{	an x} \, dx = -\int \frac{\sec^2 y}{	an y} \, dy$Let $u = 	an x$,then $du = \sec^2 x \, dx$. Let $v = 	an y$,then $dv = \sec^2 y \, dy$:$\ln |	an x| = -\ln |	an y| + C$$\ln |	an x| + \ln |	an y| = C$$\ln |	an x 	an y| = C$$	an x 	an y = e^C = K$Using the initial condition $y(\frac{\pi}{4}) = \frac{\pi}{3}$:$	an(\frac{\pi}{4}) 	an(\frac{\pi}{3}) = K$$1 	imes \sqrt{3} = K \implies K = \sqrt{3}$Thus,the solution is $	an x 	an y = \sqrt{3}$.

The solution of the differential equation $\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$ with the initial condition $y(\frac{\pi}{4}) = \frac{\pi}{3}$ is:

Similar Questions

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

The solution of the equation $\sin^{-1} \left( \frac{dy}{dx} \right) = x + y$ is

Find the general solution of the differential equation $\frac{dy}{dx} = \frac{x+1}{2-y}, (y \neq 2)$.

General solution of the differential equation $(y^3+y)(x^2+1) dy = (xy^4+2y^2x) dx$ is (where $C$ is a constant of integration.)

The particular solution of the differential equation $(1+y^2) dx - xy dy = 0$ at $x=1, y=0$,represents

Vedclass Test Series

Exam Paper Generator

Online Exam Module