The general solution of the differential equa | vedclass

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(C) Given the differential equation: $(3x-4y)(dx-3dy)+(6dx-4dy)=0$ Rearranging the terms: $(3x-4y+6)dx - (3(3x-4y)+4)dy = 0$ $\frac{dy}{dx} = \frac{3x

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$5x-15y+14\log |15x-20y-12|=c$

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(C) Given the differential equation: $(3x-4y)(dx-3dy)+(6dx-4dy)=0$ Rearranging the terms: $(3x-4y+6)dx - (3(3x-4y)+4)dy = 0$ $\frac{dy}{dx} = \frac{3x-4y+6}{3(3x-4y)+4}$ Let $t = 3x-4y$, then $\frac{dt}{dx} = 3-4\frac{dy}{dx}$, which implies $\frac{dy}{dx} = \frac{1}{4}(3-\frac{dt}{dx})$ Substituting into the equation: $\frac{1}{4}(3-\frac{dt}{dx}) = \frac{t+6}{3t+4}$ $3-\frac{dt}{dx} = \frac{4t+24}{3t+4} \Rightarrow \frac{dt}{dx} = 3 - \frac{4t+24}{3t+4} = \frac{9t+12-4t-24}{3t+4} = \frac{5t-12}{3t+4}$ Separating variables: $dx = \frac{3t+4}{5t-12}dt$ $dx = (\frac{3}{5} + \frac{56/5}{5t-12})dt$ Integrating both sides: $x = \frac{3}{5}t + \frac{56}{25}\log|5t-12| + C$ Substitute $t = 3x-4y$: $x = \frac{3}{5}(3x-4y) + \frac{56}{25}\log|5(3x-4y)-12| + C$ $x = \frac{9x-12y}{5} + \frac{56}{25}\log|15x-20y-12| + C$ $25x = 45x-60y + 56\log|15x-20y-12| + 25C$ $60y-20x = 56\log|15x-20y-12| + C'$ Dividing by $4$: $15y-5x = 14\log|15x-20y-12| + C''$ $5x-15y+14\log|15x-20y-12|=C$.

The general solution of the differential equation $(3x-4y)(dx-3dy)+(6dx-4dy)=0$ is

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