On solving frac{dy}{dx} = frac{x-y+3}{2x-2y+5 | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{x-y+3}{2(x-y)+5}$.Let $x-y = u$. Then $1 - \frac{dy}{dx} = \frac{du}{dx}$,which implies $\

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$x-y+2$

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{x-y+3}{2(x-y)+5}$.Let $x-y = u$. Then $1 - \frac{dy}{dx} = \frac{du}{dx}$,which implies $\frac{dy}{dx} = 1 - \frac{du}{dx}$.Substituting this into the equation: $1 - \frac{du}{dx} = \frac{u+3}{2u+5}$.Rearranging the terms: $\frac{du}{dx} = 1 - \frac{u+3}{2u+5} = \frac{2u+5-u-3}{2u+5} = \frac{u+2}{2u+5}$.Separating the variables: $\int \frac{2u+5}{u+2} du = \int dx$.Performing the division: $\int (2 + \frac{1}{u+2}) du = x + C$.Integrating: $2u + \log|u+2| = x + C$.Substituting $u = x-y$ back: $2(x-y) + \log|x-y+2| + C = x$.Comparing this with the given form $x = 2(x-y) + \log(t) + c$,we identify $t = x-y+2$.

On solving $\frac{dy}{dx} = \frac{x-y+3}{2x-2y+5}$,the solution obtained is $x = 2(x-y) + \log(t) + c$,find $t$.

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