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Question Number 85050 by john santu last updated on 18/Mar/20

$$(\mathrm{x}-4\mathrm{y}+3)\mathrm{dx}=(\mathrm{x}-5\mathrm{y}+4)\mathrm{dy}$$

Answered by mr W last updated on 18/Mar/20

$$let$$$$x=u+a$$$$y=v+b$$$$x-4y+3=u-4v+(a-4b+3)$$$$x-5y+4=u-5v+(a-5b+4)$$$$set$$$$a-4b+3=0$$$$a-5b+4=0$$$$\Rightarrow a=1,b=1$$$$\Rightarrow x=u+1\Rightarrow dx=du$$$$\Rightarrow y=v+1\Rightarrow dy=dv$$$$$$$$(u-4v)du=(u-5v)dv$$$$\Rightarrow \frac{dv}{du}=\frac{u-4v}{u-5v}$$$$letv=ut\Rightarrow t=\frac{v}{u}$$$$\frac{dv}{du}=t+u\frac{dt}{du}$$$$\Rightarrow t+u\frac{dt}{du}=\frac{1-4t}{1-5t}$$$$\Rightarrow u\frac{dt}{du}=\frac{5t^2-5t+1}{1-5t}$$$$\Rightarrow \frac{(1-5t)dt}{5t^2-5t+1}=\frac{du}{u}$$$$\Rightarrow \int \frac{(1-5t)dt}{5t^2-5t+1}=\int \frac{du}{u}$$$$\Rightarrow \int (\frac{1}{5t^2-5t+1}-\frac{5t}{5t^2-5t+1})dt=\ln \left(cu\right)$$$$......$$

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