The general solution of the differential equa | vedclass

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(B) Given the differential equation $\frac{dy}{dx} = \frac{2x-3y+5}{6x-9y+7}$.Let $v = 2x-3y$. Then $\frac{dv}{dx} = 2 - 3\frac{dy}{dx}$,so $\frac{dy}

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$x-3y+\frac{8}{3} \log |6x-9y-1|+c=0$

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(B) Given the differential equation $\frac{dy}{dx} = \frac{2x-3y+5}{6x-9y+7}$.Let $v = 2x-3y$. Then $\frac{dv}{dx} = 2 - 3\frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{1}{3}(2 - \frac{dv}{dx})$.Substituting into the equation: $\frac{1}{3}(2 - \frac{dv}{dx}) = \frac{v+5}{3v+7}$.$2 - \frac{dv}{dx} = \frac{3v+15}{3v+7} \implies \frac{dv}{dx} = 2 - \frac{3v+15}{3v+7} = \frac{6v+14-3v-15}{3v+7} = \frac{3v-1}{3v+7}$.Separating variables: $\int \frac{3v+7}{3v-1} dv = \int dx$.$\int (1 + \frac{8}{3v-1}) dv = \int dx \implies v + \frac{8}{3} \log |3v-1| = x + c$.Substituting $v = 2x-3y$: $(2x-3y) + \frac{8}{3} \log |3(2x-3y)-1| = x + c$.$x - 3y + \frac{8}{3} \log |6x-9y-1| + c = 0$.

The general solution of the differential equation $\frac{dy}{dx} = \frac{2x-3y+5}{6x-9y+7}$ is

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