The solution of the differential equation (1 | vedclass

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(C) Given the differential equation: $(1 + x^2)(1 + y)dy + (1 + x)(1 + y^2)dx = 0$Separate the variables:$\frac{1 + y}{1 + y^2}dy = -\frac{1 + x}{1 +

Vedclass · Accepted Answer

$	an^{-1}x + \frac{1}{2}\log(1 + x^2) + 	an^{-1}y + \frac{1}{2}\log(1 + y^2) = c$

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(C) Given the differential equation: $(1 + x^2)(1 + y)dy + (1 + x)(1 + y^2)dx = 0$Separate the variables:$\frac{1 + y}{1 + y^2}dy = -\frac{1 + x}{1 + x^2}dx$Integrate both sides:$\int \frac{1 + y}{1 + y^2}dy = -\int \frac{1 + x}{1 + x^2}dx$Split the integrals:$\int \left( \frac{1}{1 + y^2} + \frac{y}{1 + y^2} ight)dy = -\int \left( \frac{1}{1 + x^2} + \frac{x}{1 + x^2} ight)dx$Perform the integration:$	an^{-1}y + \frac{1}{2}\log(1 + y^2) = -(	an^{-1}x + \frac{1}{2}\log(1 + x^2)) + C$Rearrange the terms to get the final solution:$	an^{-1}x + \frac{1}{2}\log(1 + x^2) + 	an^{-1}y + \frac{1}{2}\log(1 + y^2) = C$

The solution of the differential equation $(1 + x^2)(1 + y)dy + (1 + x)(1 + y^2)dx = 0$ is

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