General solution of the differential equation | vedclass

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(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+

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

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(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

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