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(C) Let $x - y = v$. Then,differentiating with respect to $x$,we get $1 - \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{

Vedclass · Accepted Answer

$2(x - y) + \log(x - y + 2) = x + c$

Vedclass · Answer

(C) Let $x - y = v$. Then,differentiating with respect to $x$,we get $1 - \frac{dy}{dx} = \frac{dv}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{dv}{dx}$.Substituting this into the given differential equation: $1 - \frac{dv}{dx} = \frac{v + 3}{2v + 5}$.Rearranging the terms: $\frac{dv}{dx} = 1 - \frac{v + 3}{2v + 5} = \frac{2v + 5 - v - 3}{2v + 5} = \frac{v + 2}{2v + 5}$.Separating the variables: $\int \frac{2v + 5}{v + 2} dv = \int dx$.Simplifying the integrand: $\int \left( 2 + \frac{1}{v + 2} \right) dv = \int dx$.Integrating both sides: $2v + \log|v + 2| = x + c$.Substituting $v = x - y$ back into the equation: $2(x - y) + \log|x - y + 2| = x + c$.

The solution of the differential equation $\frac{dy}{dx} = \frac{x - y + 3}{2(x - y) + 5}$ is

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