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Question Number 15095 by tawa tawa last updated on 07/Jun/17

Solve:  (1 − x)(dy/dx) = y(1 + x)

$$\mathrm{Solve}: \\ $$$$\left(\mathrm{1}\:−\:\mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{y}\left(\mathrm{1}\:+\:\mathrm{x}\right) \\ $$

Answered by Tinkutara last updated on 07/Jun/17

(dy/y) = ((1 + x)/(1 − x)) dx = (−1 − (2/(x − 1)))dx  ln y = −x − 2 ln ∣x − 1∣ + C  ln y + ln (x − 1)^2  = C − x  ln y(x − 1)^2  = C − x  (e^C /e^x ) = y(x − 1)^2   y(x − 1)^2 e^x  = C

$$\frac{{dy}}{{y}}\:=\:\frac{\mathrm{1}\:+\:{x}}{\mathrm{1}\:−\:{x}}\:{dx}\:=\:\left(−\mathrm{1}\:−\:\frac{\mathrm{2}}{{x}\:−\:\mathrm{1}}\right){dx} \\ $$$$\mathrm{ln}\:{y}\:=\:−{x}\:−\:\mathrm{2}\:\mathrm{ln}\:\mid{x}\:−\:\mathrm{1}\mid\:+\:{C} \\ $$$$\mathrm{ln}\:{y}\:+\:\mathrm{ln}\:\left({x}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:{C}\:−\:{x} \\ $$$$\mathrm{ln}\:{y}\left({x}\:−\:\mathrm{1}\right)^{\mathrm{2}} \:=\:{C}\:−\:{x} \\ $$$$\frac{{e}^{{C}} }{{e}^{{x}} }\:=\:{y}\left({x}\:−\:\mathrm{1}\right)^{\mathrm{2}} \\ $$$${y}\left({x}\:−\:\mathrm{1}\right)^{\mathrm{2}} {e}^{{x}} \:=\:{C} \\ $$

Commented by tawa tawa last updated on 07/Jun/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

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