The general solution of frac{dy}{dx} = frac{x | vedclass

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Question

(A) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1} \dots (i)$ Let $x+y = v$. Then,differentiating with respect to $x$,we get $1

Vedclass · Accepted Answer

$y = x + \log(x+y) + c$,where $c$ is a constant of integration.

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{x+y+1}{x+y-1} \dots (i)$ Let $x+y = v$. Then,differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = \frac{dv}{dx} - 1 \dots (ii)$ Substituting $(ii)$ into $(i)$: $\frac{dv}{dx} - 1 = \frac{v+1}{v-1}$ $\frac{dv}{dx} = \frac{v+1}{v-1} + 1 = \frac{v+1+v-1}{v-1} = \frac{2v}{v-1}$ Separating the variables: $\frac{v-1}{2v} dv = dx$ $\frac{1}{2} (1 - \frac{1}{v}) dv = dx$ Integrating both sides: $\frac{1}{2} (v - \log|v|) = x + c_1$ $v - \log|v| = 2x + 2c_1$ Substituting $v = x+y$: $(x+y) - \log|x+y| = 2x + c$ $y - \log|x+y| = x + c$ $y = x + \log|x+y| + c$,where $c = -2c_1$.

The general solution of $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$ is

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