The general solution of the differential equa | vedclass

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(C) Given the differential equation $\frac{dy}{dx} = \frac{x+2y-1}{x+2y+1}$.Let $t = x+2y$. Then $\frac{dt}{dx} = 1 + 2\frac{dy}{dx}$,which implies $\

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$6(-x+y)+4 \log |3x+6y-1| = K$

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(C) Given the differential equation $\frac{dy}{dx} = \frac{x+2y-1}{x+2y+1}$.Let $t = x+2y$. Then $\frac{dt}{dx} = 1 + 2\frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2}\left(\frac{dt}{dx} - 1\right)$.Substituting this into the equation: $\frac{1}{2}\left(\frac{dt}{dx} - 1\right) = \frac{t-1}{t+1}$.$\frac{dt}{dx} - 1 = \frac{2t-2}{t+1} \Rightarrow \frac{dt}{dx} = \frac{2t-2}{t+1} + 1 = \frac{2t-2+t+1}{t+1} = \frac{3t-1}{t+1}$.Separating the variables: $\int \frac{t+1}{3t-1} dt = \int dx$.To integrate $\frac{t+1}{3t-1}$,we write it as $\frac{1}{3} \int \frac{3t+3}{3t-1} dt = \frac{1}{3} \int \frac{3t-1+4}{3t-1} dt = \frac{1}{3} \int (1 + \frac{4}{3t-1}) dt$.This gives $\frac{1}{3} (t + \frac{4}{3} \log |3t-1|) = x + C$.Multiplying by $9$: $3t + 4 \log |3t-1| = 9x + 9C$.Substituting $t = x+2y$: $3(x+2y) + 4 \log |3(x+2y)-1| = 9x + K$.$3x + 6y + 4 \log |3x+6y-1| = 9x + K$.$6y - 6x + 4 \log |3x+6y-1| = K$.$6(-x+y) + 4 \log |3x+6y-1| = K$.

The general solution of the differential equation $\frac{dy}{dx} = \frac{x+2y-1}{x+2y+1}$ is

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