The solution of the differential equation (1 | vedclass

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

Question

(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$

Vedclass · Answer

(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

Similar Questions

The general solution of the differential equation $(9x - 3y + 5) dy = (3x - y + 1) dx$ is

The particular solution of the differential equation $\log\left(\frac{dy}{dx}\right) = x$,when $x = 0, y = 1$ is .....

The solution of $(x - y^3)dx + 3xy^2dy = 0$ is

The solution of the differential equation $(1 + x^2)\frac{dy}{dx} = x(1 + y^2)$ is

The solution of $x^2 + y^2 \frac{dy}{dx} = 4$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module