The general solution of the differential equa | vedclass

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(B) Given differential equation: $(9x - 3y + 5) dy = (3x - y + 1) dx$ $\frac{dy}{dx} = \frac{3x - y + 1}{3(3x - y) + 5}$ Let $v = 3x - y$. Then $\frac

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$4x - 12y - \log |12x - 4y + 7| = c$

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(B) Given differential equation: $(9x - 3y + 5) dy = (3x - y + 1) dx$ $\frac{dy}{dx} = \frac{3x - y + 1}{3(3x - y) + 5}$ Let $v = 3x - y$. Then $\frac{dv}{dx} = 3 - \frac{dy}{dx}$,so $\frac{dy}{dx} = 3 - \frac{dv}{dx}$. Substituting into the equation: $3 - \frac{dv}{dx} = \frac{v + 1}{3v + 5}$ $\frac{dv}{dx} = 3 - \frac{v + 1}{3v + 5} = \frac{9v + 15 - v - 1}{3v + 5} = \frac{8v + 14}{3v + 5}$ Separating variables: $\frac{3v + 5}{8v + 14} dv = dx$ $\frac{1}{8} \int \frac{3v + 5}{v + 1.75} dv = x + C$ Using partial fractions or adjustment: $\frac{3v + 5}{8v + 14} = \frac{3}{8} \left( \frac{8v + 14 - 14 + 13.33}{8v + 14} \right) = \frac{3}{8} \left( 1 + \frac{-14/3 + 5}{8v + 14} \right) = \frac{3}{8} - \frac{1}{8(8v + 14)} \dots$ Integrating: $\frac{3}{8} v - \frac{1}{64} \ln |8v + 14| = x + C$ Substitute $v = 3x - y$: $\frac{3}{8}(3x - y) - \frac{1}{64} \ln |8(3x - y) + 14| = x + C$ Multiply by $64$: $24(3x - y) - \ln |24x - 8y + 14| = 64x + C'$ $72x - 24y - \ln |2(12x - 4y + 7)| = 64x + C'$ $8x - 24y - \ln |12x - 4y + 7| = C''$ Dividing by $2$: $4x - 12y - \frac{1}{2} \ln |12x - 4y + 7| = C$. This matches option $B$.

The general solution of the differential equation $(9x - 3y + 5) dy = (3x - y + 1) dx$ is

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