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

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(A) Given the differential equation $\frac{dy}{dx} = \frac{x+y+1}{y-x+1}$.Cross-multiplying, we get $(y-x+1) dy = (x+y+1) dx$.Rearranging the terms: $

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$2xy + (x+1)^2 - (y+1)^2 = C$

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(A) Given the differential equation $\frac{dy}{dx} = \frac{x+y+1}{y-x+1}$.Cross-multiplying, we get $(y-x+1) dy = (x+y+1) dx$.Rearranging the terms: $y dy - x dy + dy = x dx + y dx + dx$.Grouping terms: $(y+1) dy - x dy = (x+1) dx + y dx$.Adding $x dy$ to both sides: $(y+1) dy = (x+1) dx + (y dx + x dy)$.Recognizing the product rule $d(xy) = y dx + x dy$, we have $(y+1) dy = (x+1) dx + d(xy)$.Integrating both sides: $\int (y+1) dy = \int (x+1) dx + \int d(xy)$.This yields $\frac{(y+1)^2}{2} = \frac{(x+1)^2}{2} + xy + C_1$.Multiplying by $2$: $(y+1)^2 = (x+1)^2 + 2xy + 2C_1$.Rearranging: $2xy + (x+1)^2 - (y+1)^2 = -2C_1$.Letting $C = -2C_1$, we get $2xy + (x+1)^2 - (y+1)^2 = C$.

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

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