If 2x - y + c log(x - 2y - 4) = k is the gene | vedclass

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(C) Given the differential equation $\frac{dy}{dx} = \frac{2x - 4y - 5}{x - 2y + 2}$.We can rewrite the numerator as $2(x - 2y + 2) - 9$.So,$\frac{dy}

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$3$

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(C) Given the differential equation $\frac{dy}{dx} = \frac{2x - 4y - 5}{x - 2y + 2}$.We can rewrite the numerator as $2(x - 2y + 2) - 9$.So,$\frac{dy}{dx} = \frac{2(x - 2y + 2) - 9}{x - 2y + 2}$.Let $t = x - 2y + 2$. Then $\frac{dt}{dx} = 1 - 2\frac{dy}{dx}$,which implies $\frac{dy}{dx} = \frac{1}{2}(1 - \frac{dt}{dx})$.Substituting this into the equation: $\frac{1}{2}(1 - \frac{dt}{dx}) = \frac{2t - 9}{t} = 2 - \frac{9}{t}$.$1 - \frac{dt}{dx} = 4 - \frac{18}{t} \Rightarrow \frac{dt}{dx} = \frac{18}{t} - 3 = \frac{18 - 3t}{t}$.Separating variables: $\frac{t}{18 - 3t} dt = dx \Rightarrow \frac{t}{3(6 - t)} dt = dx$.Using partial fractions: $-\frac{1}{3} \int \frac{t - 6 + 6}{t - 6} dt = \int dx \Rightarrow -\frac{1}{3} \int (1 + \frac{6}{t - 6}) dt = x + C_1$.$-\frac{1}{3} (t + 6 \ln|t - 6|) = x + C_1 \Rightarrow -t - 6 \ln|t - 6| = 3x + C$.Substituting $t = x - 2y + 2$: $-(x - 2y + 2) - 6 \ln|x - 2y + 2 - 6| = 3x + C$.$-x + 2y - 2 - 6 \ln|x - 2y - 4| = 3x + C \Rightarrow -4x + 2y - 6 \ln|x - 2y - 4| = C + 2$.Dividing by $-2$: $2x - y + 3 \ln|x - 2y - 4| = k$.Comparing with $2x - y + c \log(x - 2y - 4) = k$,we get $c = 3$.

If $2x - y + c \log(x - 2y - 4) = k$ is the general solution of $\frac{dy}{dx} = \frac{2x - 4y - 5}{x - 2y + 2}$,then $c =$

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