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Question Number 159319 by nimnim last updated on 15/Nov/21

$$\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{x}-\mathrm{y}-2}{\mathrm{x}-2\mathrm{y}-3}=0$$$$\mathrm{Pls}...\mathrm{solve}\mathrm{the}\mathrm{differential}\mathrm{equation}.$$

Answered by mr W last updated on 15/Nov/21

$$let$$$$x=u+a$$$$y=v+b$$$$x-y-2=u-v+(a-b-2)$$$$x-2y-3=u-2v+(a-2b-3)$$$$let$$$$a-b-2=0$$$$a-2b-3=0$$$$\Rightarrow a=1,b=-1$$$$x=u+1,y=v-1$$$$dx=du,dy=dv$$$$\frac{dv}{du}+\frac{u-v}{u-2v}=0$$$$letv=ut$$$$\frac{dv}{du}=t+u\frac{dt}{du}$$$$t+u\frac{dt}{du}+\frac{1-t}{1-2t}=0$$$$u\frac{dt}{du}=\frac{t-1}{1-2t}-t=\frac{2t^2-1}{1-2t}$$$$\frac{(1-2t)dt}{2t^2-1}=\frac{du}{du}$$$$\int \frac{(1-2t)dt}{2t^2-1}=\int \frac{du}{du}$$$$\frac{1}{2\sqrt{2}}\ln \frac{2t-\sqrt{2}}{2t+\sqrt{2}}-\frac{1}{2}\ln (2t^2-1)=\ln u+C$$$$\frac{1}{2\sqrt{2}}\ln \frac{\frac{2v}{u}-\sqrt{2}}{\frac{2v}{u}+\sqrt{2}}-\frac{1}{2}\ln (\frac{2v^2}{u^2}-1)=\ln u+C$$$$\ln \frac{\frac{2(y+1)}{x-1}-\sqrt{2}}{\frac{2(y+1)}{x-1}+\sqrt{2}}-\sqrt{2}\ln [\frac{2{(y+1)}^2}{{(x-1)}^2}-1]=2\sqrt{2}\ln (x-1)+C$$

Commented by nimnim last updated on 16/Nov/21

$$\mathrm{thank}\mathrm{you}\mathrm{Sir}..$$

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