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Question Number 116329 by bemath last updated on 03/Oct/20

$$\mathrm{solve}\mathrm{the}\mathrm{diff}\mathrm{equation}$$$$\left(1\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}+3\mathrm{y}-5}{\mathrm{x}-\mathrm{y}-1}$$$$\left(2\right)(3\mathrm{y}-7\mathrm{x}-3)\mathrm{dx}+(7\mathrm{y}-3\mathrm{x}-7)\mathrm{dy}=0$$

Answered by mr W last updated on 03/Oct/20

$$\left(1\right)\frac{dy}{dx}=\frac{x+3y-5}{x-y-1}$$$$let$$$$x=u+a$$$$y=v+b$$$$\Rightarrow x+3y-5=u+3v+a+3b-5$$$$\Rightarrow x-y-1=u-v+a-b-1$$$$a+3b-5=0...\left(i\right)$$$$a-b-1=0...\left(ii\right)$$$$\Rightarrow a=2$$$$\Rightarrow b=1$$$$\Rightarrow x=u+2$$$$\Rightarrow y=v+1$$$$\frac{dy}{dx}=\frac{dv}{du}=\frac{u+3v}{u-v}$$$$v=tu$$$$\frac{dv}{du}=t+u\frac{dt}{du}=\frac{1+3t}{1-t}$$$$u\frac{dt}{du}=\frac{1+3t}{1-t}-t=\frac{{(1+t)}^2}{1-t}$$$$\Rightarrow \frac{(1-t)dt}{{(1+t)}^2}=\frac{du}{u}$$$$\Rightarrow [\frac{2}{{(1+t)}^2}-\frac{1}{1+t}]dt=\frac{du}{u}$$$$\Rightarrow -\frac{2}{1+t}-\ln (1+t)=\ln u+C$$$$\Rightarrow \ln u(1+t)=-(\frac{2}{1+t}+C)$$$$\Rightarrow u(1+t)={Ce}^{-\frac{2}{1+t}}$$$$\Rightarrow (x-2)(1+\frac{y-1}{x-2})={Ce}^{-\frac{2}{1+\frac{y-1}{x-2}}}$$$$\Rightarrow x+y-3={Ce}^{-\frac{2(x-2)}{x+y-3}}$$$$\Rightarrow C(x-2)=\frac{-2(x-2)}{x+y-3}{e}^{-\frac{2(x-2)}{x+y-3}}$$$$\Rightarrow \frac{-2(x-2)}{x+y-3}=\mathbb{W}\left(C(x-2)\right)$$$$\Rightarrow y=3-\frac{2(x-2)}{\mathbb{W}\left(C(x-2)\right)}-x$$

Commented by bemath last updated on 03/Oct/20

$$\mathrm{W}\mathrm{is}\mathrm{Lambert}\mathrm{W}\mathrm{Function}?\mathrm{thankyou}$$$$\mathrm{sir}$$

Commented by mr W last updated on 03/Oct/20

$$yes,WforlambertWfunction!$$

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