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Question Number 31485 by mondodotto@gmail.com last updated on 09/Mar/18

Answered by ajfour last updated on 09/Mar/18

$$\left(i\right)lety-2=t(x-3)$$$$\Rightarrow \frac{dy}{dx}=(x-3)\frac{dt}{dx}+t$$$$Thendiff.eq.becomes$$$$(x-3)\frac{dt}{dx}+t=\frac{t(x-3)}{(t+1)(x-3)}$$$$\Rightarrow (x-3)\frac{dt}{dx}+t=\frac{t}{t+1}$$$$or(x-3)\frac{dt}{dx}=\frac{t}{t+1}-t$$$$\int \frac{(t+1)dt}{t^2}=-\int \frac{dx}{x-3}$$$$\Rightarrow \ln \mid t\mid -\frac{1}{t}=-\ln \mid x-3\mid +c$$$$\Rightarrow \ln \mid \frac{y-2}{x-3}\mid +\ln \mid x-3\mid =\frac{x-3}{y-2}+c$$$$\Rightarrow \ln \mid y-2\mid =\frac{x-3}{y-2}+c.$$

Answered by ajfour last updated on 09/Mar/18

$$\left(ii\right)let2x+y+1=t$$$$\Rightarrow \frac{dy}{dx}=\frac{dt}{dx}-2$$$$Differentialeq.becomes$$$$\frac{dt}{dx}-2=\frac{t-3}{t}$$$$\Rightarrow \frac{dt}{dx}=\frac{3t-3}{t}$$$$or\int \frac{t}{t-1}dt=3\int dx$$$$\Rightarrow \int dt+\int \frac{dt}{t-1}=3x+c$$$$ort+\ln \mid t-1\mid =3x+c$$$$\Rightarrow 2x+y+1+\ln \mid 2x+y\mid =3x+c.$$

Commented by mondodotto@gmail.com last updated on 09/Mar/18

$$\mathrm{thanx}\mathrm{a}\mathrm{lot}.$$

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