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: $\frac{\co

Vedclass · Accepted Answer

$(1+\sin x)(1+\cos y) = c$,where $c$ is a constant of integration.

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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: $\frac{\cos x}{1+\sin x} dx = \frac{\sin y}{1+\cos y} dy$. 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 $dv = -\sin y dy$. Substituting these into the integral: $\int \frac{1}{u} du = -\int \frac{1}{v} dv$. $\ln |u| = -\ln |v| + \ln |c|$. $\ln |u| + \ln |v| = \ln |c|$. $\ln |uv| = \ln |c|$. $uv = c$. Substituting back the values of $u$ and $v$: $(1+\sin x)(1+\cos y) = c$.

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

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