The solution of the differential equation fra | vedclass

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

Question

(C) Given the differential equation: $\frac{dy}{dx} = \frac{(1+x)y}{(y-1)x}$.Separating the variables,we get:$\frac{y-1}{y} dy = \frac{1+x}{x} dx$Inte

Vedclass · Accepted Answer

$\log(xy) + x - y = c$

Vedclass · Answer

(C) Given the differential equation: $\frac{dy}{dx} = \frac{(1+x)y}{(y-1)x}$.Separating the variables,we get:$\frac{y-1}{y} dy = \frac{1+x}{x} dx$Integrating both sides:$\int (1 - \frac{1}{y}) dy = \int (1 + \frac{1}{x}) dx$$y - \log|y| = x + \log|x| + c$Rearranging the terms:$y - x - c = \log|x| + \log|y|$$y - x - c = \log|xy|$$\log|xy| + x - y = -c$Since $-c$ is also an arbitrary constant,we can write it as $C$:$\log|xy| + x - y = C$.

The solution of the differential equation $\frac{dy}{dx} = \frac{(1+x)y}{(y-1)x}$ is (where $c$ is the constant of integration):

Similar Questions

The solution of $\frac{dy}{dx} = (\frac{x}{y})^{-1/3}$ is

The solution of the differential equation $\frac{dy}{dx} = \frac{x-2y+1}{2x-4y}$ is

The particular solution of the differential equation $\frac{dy}{dx} = \sec y$ with the initial condition $y(0) = 0$ is:

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

Find the equation of a curve passing through the point $(-2, 3),$ given that the slope of the tangent to the curve at any point $(x, y)$ is $\frac{2x}{y^2}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module