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 solution of $\frac{dy}{dx} = \frac{e^x(\sin^2 x + \sin 2x)}{y(2\log y + 1)}$ is

If $x^{3} dy + xy dx = x^{2} dy + 2y dx$,$y(2) = e$ and $x > 1$,then $y(4)$ is equal to

If $y = y(x)$ is the solution of the differential equation $2x^{2} \frac{dy}{dx} - 2xy + 3y^{2} = 0$ such that $y(e) = \frac{e}{3}$,then $y(1)$ is equal to

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

The general solution $y(x)$ of the differential equation $\frac{d}{dy} \left( \int_x^y dt \right) = x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module