General solution of the differential equation | vedclass

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Question

(C) Given the differential equation $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.Let $x+y = t$.Differentiating with respect to $x$,we get $1 + \frac{dy}{dx}

Vedclass · Accepted Answer

$y = x + \log |x+y| + c$

Vedclass · Answer

(C) Given the differential equation $\frac{dy}{dx} = \frac{x+y+1}{x+y-1}$.Let $x+y = t$.Differentiating with respect to $x$,we get $1 + \frac{dy}{dx} = \frac{dt}{dx}$,which implies $\frac{dy}{dx} = \frac{dt}{dx} - 1$.Substituting this into the original equation:$\frac{dt}{dx} - 1 = \frac{t+1}{t-1}$$\frac{dt}{dx} = \frac{t+1}{t-1} + 1 = \frac{t+1+t-1}{t-1} = \frac{2t}{t-1}$.Separating the variables,we get $\left(\frac{t-1}{2t}\right) dt = dx$.This simplifies to $\left(\frac{1}{2} - \frac{1}{2t}\right) dt = dx$.Integrating both sides:$\int \left(\frac{1}{2} - \frac{1}{2t}\right) dt = \int dx$$\frac{1}{2}t - \frac{1}{2} \log |t| = x + C_1$.Multiplying by $2$:$t - \log |t| = 2x + 2C_1$.Substituting $t = x+y$ back:$(x+y) - \log |x+y| = 2x + C$ (where $C = 2C_1$).$y - x = \log |x+y| + C$,which can be rearranged as $y = x + \log |x+y| + C$.

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

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