General solution of the differential equation | vedclass

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

Question

(C) Given differential equation is $\cos x(1+\cos y) dx - \sin y(1+\sin x) dy = 0$. Rearranging the terms to separate the variables,we get: $\cos x(1+

Vedclass · Accepted Answer

$(1+\sin x)(1+\cos y) = c$

Vedclass · Answer

(C) Given differential equation is $\cos x(1+\cos y) dx - \sin y(1+\sin x) dy = 0$. Rearranging the terms to separate the variables,we get: $\cos x(1+\cos y) dx = \sin y(1+\sin x) dy$ $\frac{\cos x}{1+\sin x} dx = \frac{\sin y}{1+\cos y} dy$ Integrating both sides: $\int \frac{\cos x}{1+\sin x} dx = \int \frac{\sin y}{1+\cos y} dy$ Let $u = 1+\sin x$,then $du = \cos x dx$. Let $v = 1+\cos y$,then $dv = -\sin y dy$,so $\sin y dy = -dv$. Substituting these into the integral: $\int \frac{1}{u} du = \int -\frac{1}{v} dv$ $\ln|u| = -\ln|v| + \ln|c|$ $\ln|1+\sin x| = -\ln|1+\cos y| + \ln|c|$ $\ln|1+\sin x| + \ln|1+\cos y| = \ln|c|$ $\ln|(1+\sin x)(1+\cos y)| = \ln|c|$ Taking the exponential of both sides: $(1+\sin x)(1+\cos y) = c$.

General solution of the differential equation $\cos x(1+\cos y) dx - \sin y(1+\sin x) dy = 0$ is

Similar Questions

The particular solution of the differential equation $y(\frac{dx}{dy}) = x \log x$ at $x = e$ and $y = 1$ is

The solution of the differential equation $\cos(x+y) dy = dx$ given that $y(0) = 0$ is

Find the particular solution of the differential equation $\frac{dx}{dy} = \frac{\sin y(1 + y \cot y)}{x \log(x^2 e)}$,given that $y(1) = 0$.

The general solution of ${x^2}\frac{{dy}}{{dx}} = 2$ is

The particular solution of the differential equation $x \, dy + 2y \, dx = 0$, given that $y = 1$ when $x = 2$, is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module