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: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.Let $u = x+y$. Then,differentiating with respect to $x$,we get $1 + \frac{d

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.Let $u = x+y$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{du}{dx}$,which implies $\frac{dy}{dx} = \frac{du}{dx} - 1$.Substituting this into the equation:$\frac{du}{dx} - 1 = \frac{u+1}{u-1}$$\frac{du}{dx} = \frac{u+1}{u-1} + 1 = \frac{u+1+u-1}{u-1} = \frac{2u}{u-1}$.Separating the variables:$\left(\frac{u-1}{u}\right) du = 2 dx$$(1 - \frac{1}{u}) du = 2 dx$.Integrating both sides:$\int (1 - \frac{1}{u}) du = \int 2 dx$$u - \log|u| = 2x + c$.Substituting $u = x+y$ back:$(x+y) - \log|x+y| = 2x + c$$y - x = \log|x+y| + c$ or $y = x + \log|x+y| + c$.

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

Similar Questions

The solution of the differential equation $x \sec y \frac{dy}{dx} = 1$ is

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

The particular solution of the differential equation $\frac{dy}{dx} = -4xy^2$ with the initial condition $x = 0, y = 1$ is . . . . . . .

The general solution of the differential equation ${e^y}\frac{{dy}}{{dx}} + ({e^y} + 1)\cot x = 0$ is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module