The general solution of the differential equa | vedclass

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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

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