The general solution of the differential equa | vedclass

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

Question

(D) Given the differential equation: $\frac{dy}{dx} = \frac{xy+x-2y-2}{xy-2x+y-2}$ Factorizing the numerator and denominator: $\frac{dy}{dx} = \frac{x

Vedclass · Accepted Answer

$x-y+3 \log \left|\frac{y+1}{x+1}\right|=c$

Vedclass · Answer

(D) Given the differential equation: $\frac{dy}{dx} = \frac{xy+x-2y-2}{xy-2x+y-2}$ Factorizing the numerator and denominator: $\frac{dy}{dx} = \frac{x(y+1)-2(y+1)}{x(y-2)+1(y-2)} = \frac{(x-2)(y+1)}{(x+1)(y-2)}$ Separating the variables: $\frac{y-2}{y+1} dy = \frac{x-2}{x+1} dx$ Rewrite the fractions: $\frac{y+1-3}{y+1} dy = \frac{x+1-3}{x+1} dx$ $(1 - \frac{3}{y+1}) dy = (1 - \frac{3}{x+1}) dx$ Integrating both sides: $\int (1 - \frac{3}{y+1}) dy = \int (1 - \frac{3}{x+1}) dx$ $y - 3 \ln |y+1| = x - 3 \ln |x+1| + c$ Rearranging the terms: $x - y + 3 \ln |y+1| - 3 \ln |x+1| = c$ $x - y + 3 \ln \left| \frac{y+1}{x+1} \right| = c$

The general solution of the differential equation $\frac{dy}{dx} = \frac{xy+x-2y-2}{xy-2x+y-2}$ is

Similar Questions

The solution of the differential equation $y \ dx - x \ dy = xy \ dx$ is . . . . . .

The general solution of the differential equation $\frac{dy}{dx} = (9x + y + 5)^2$ is...

The general solution of the differential equation $\frac{dy}{dx} = \frac{1}{x+y+1}$ is ($k, c$ are arbitrary constants)

The particular solution of the differential equation $x dy + 2y dx = 0$,when $x = 2$ and $y = 1$ is

The solution of the differential equation $x \, dy + y \, dx - \sqrt{1 - x^2 y^2} \, dx = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module