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Question Number 93620 by i jagooll last updated on 14/May/20

$$(3\mathrm{x}+3\mathrm{y}-4)\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}+\mathrm{y}$$

Answered by john santu last updated on 14/May/20

$$r=x+y\Rightarrow \frac{dr}{dx}=1+\frac{dy}{dx}$$$$(3r-4)(\frac{dr}{dx}-1)=r$$$$\frac{dr}{dx}=\frac{r}{3r-4}+\frac{3r-4}{3r-4}=\frac{4r-4}{3r-4}$$$$\frac{(3r-4)}{r-1}dr=4dx$$$$\int \frac{3(r-1)-1}{r-1}dr=4x+c$$$$3r-\ln (r-1)=4x+c$$$$3x+3y-\ln (x+y-1)=4x+c$$$$3y-\ln (x+y-1)-x=C$$

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