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: $\sec^2 x 	an y \, dx + \sec^2 y 	an x \, dy = 0$. Divide both sides by $	an x 	an y$: $\frac{\sec^2 x}{	an

Vedclass · Accepted Answer

$	an x 	an y = c$

Vedclass · Answer

(C) Given the differential equation: $\sec^2 x 	an y \, dx + \sec^2 y 	an x \, dy = 0$. Divide both sides by $	an x 	an y$: $\frac{\sec^2 x}{	an x} \, dx + \frac{\sec^2 y}{	an y} \, dy = 0$. Integrating both sides: $\int \frac{\sec^2 x}{	an x} \, dx + \int \frac{\sec^2 y}{	an y} \, dy = C_1$. Let $u = 	an x$,then $du = \sec^2 x \, dx$. Let $v = 	an y$,then $dv = \sec^2 y \, dy$. Substituting these into the integral: $\int \frac{1}{u} \, du + \int \frac{1}{v} \, dv = C_1$. $\ln|u| + \ln|v| = C_1$. $\ln|	an x| + \ln|	an y| = C_1$. Using the property $\ln a + \ln b = \ln(ab)$: $\ln|	an x 	an y| = C_1$. Taking the exponential of both sides: $|	an x 	an y| = e^{C_1}$. Let $e^{C_1} = c$,so $	an x 	an y = c$.

The general solution of the differential equation $\sec^2 x \tan y \, dx + \sec^2 y \tan x \, dy = 0$ is . . . . . . .

Similar Questions

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

The solution of the differential equation $(1 - x^2)(1 - y)dx = xy(1 + y)dy$ is

If $y=y(x)$ is the solution of the differential equation $\left(\frac{5+e^x}{2+y}\right) \frac{dy}{dx}+e^x=0$ satisfying $y(0)=1$,then a value of $y(\log 13)$ is

The solution of the differential equation $({x^2} - y{x^2})\frac{{dy}}{{dx}} + {y^2} + x{y^2} = 0$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module