Question Number 159319 by nimnim last updated on 15/Nov/21
$$\:\:\:\:\:\:\:\frac{\mathrm{dy}}{\mathrm{dx}}+\frac{\mathrm{x}−\mathrm{y}−\mathrm{2}}{\mathrm{x}−\mathrm{2y}−\mathrm{3}}=\mathrm{0} \\ $$$$\mathrm{Pls}…\:\mathrm{solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}. \\ $$
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 ⇒a=1, b=−1 x=u+1, y=v−1 dx=du, dy=dv (dv/du)+((u−v)/(u−2v))=0 let v=ut (dv/du)=t+u(dt/du) t+u(dt/du)+((1−t)/(1−2t))=0 u(dt/du)=((t−1)/(1−2t))−t=((2t^2 −1)/(1−2t)) (((1−2t)dt)/(2t^2 −1))=(du/du) ∫(((1−2t)dt)/(2t^2 −1))=∫(du/du) (1/(2(√2)))ln ((2t−(√2))/(2t+(√2)))−(1/2)ln (2t^2 −1)=ln u+C (1/(2(√2)))ln ((((2v)/u)−(√2))/(((2v)/u)+(√2)))−(1/2)ln (((2v^2 )/u^2 )−1)=ln u+C ln ((((2(y+1))/(x−1))−(√2))/(((2(y+1))/(x−1))+(√2)))−(√2)ln [((2(y+1)^2 )/((x−1)^2 ))−1]=2(√2)ln (x−1)+C](https://www.tinkutara.com/question/Q159340.png)
$${let} \\ $$$${x}={u}+{a} \\ $$$${y}={v}+{b} \\ $$$${x}−{y}−\mathrm{2}={u}−{v}+\left({a}−{b}−\mathrm{2}\right) \\ $$$${x}−\mathrm{2}{y}−\mathrm{3}={u}−\mathrm{2}{v}+\left({a}−\mathrm{2}{b}−\mathrm{3}\right) \\ $$$${let}\: \\ $$$${a}−{b}−\mathrm{2}=\mathrm{0} \\ $$$${a}−\mathrm{2}{b}−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{a}=\mathrm{1},\:{b}=−\mathrm{1} \\ $$$${x}={u}+\mathrm{1},\:{y}={v}−\mathrm{1} \\ $$$${dx}={du},\:{dy}={dv} \\ $$$$\frac{{dv}}{{du}}+\frac{{u}−{v}}{{u}−\mathrm{2}{v}}=\mathrm{0} \\ $$$${let}\:{v}={ut} \\ $$$$\frac{{dv}}{{du}}={t}+{u}\frac{{dt}}{{du}} \\ $$$${t}+{u}\frac{{dt}}{{du}}+\frac{\mathrm{1}−{t}}{\mathrm{1}−\mathrm{2}{t}}=\mathrm{0} \\ $$$${u}\frac{{dt}}{{du}}=\frac{{t}−\mathrm{1}}{\mathrm{1}−\mathrm{2}{t}}−{t}=\frac{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}}{\mathrm{1}−\mathrm{2}{t}} \\ $$$$\frac{\left(\mathrm{1}−\mathrm{2}{t}\right){dt}}{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}}=\frac{{du}}{{du}} \\ $$$$\int\frac{\left(\mathrm{1}−\mathrm{2}{t}\right){dt}}{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}}=\int\frac{{du}}{{du}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\:\frac{\mathrm{2}{t}−\sqrt{\mathrm{2}}}{\mathrm{2}{t}+\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\mathrm{2}{t}^{\mathrm{2}} −\mathrm{1}\right)=\mathrm{ln}\:{u}+{C} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{ln}\:\frac{\frac{\mathrm{2}{v}}{{u}}−\sqrt{\mathrm{2}}}{\frac{\mathrm{2}{v}}{{u}}+\sqrt{\mathrm{2}}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left(\frac{\mathrm{2}{v}^{\mathrm{2}} }{{u}^{\mathrm{2}} }−\mathrm{1}\right)=\mathrm{ln}\:{u}+{C} \\ $$$$\mathrm{ln}\:\frac{\frac{\mathrm{2}\left({y}+\mathrm{1}\right)}{{x}−\mathrm{1}}−\sqrt{\mathrm{2}}}{\frac{\mathrm{2}\left({y}+\mathrm{1}\right)}{{x}−\mathrm{1}}+\sqrt{\mathrm{2}}}−\sqrt{\mathrm{2}}\mathrm{ln}\:\left[\frac{\mathrm{2}\left({y}+\mathrm{1}\right)^{\mathrm{2}} }{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }−\mathrm{1}\right]=\mathrm{2}\sqrt{\mathrm{2}}\mathrm{ln}\:\left({x}−\mathrm{1}\right)+{C} \\ $$
Commented by nimnim last updated on 16/Nov/21
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{Sir}.. \\ $$