Question Number 204180 by naka3546 last updated on 08/Feb/24
$$\mathrm{Solve}\:\:\mathrm{this}. \\ $$$$ \\ $$$$\frac{{x}+{y}−{a}}{{x}+{y}−{b}}\:\centerdot\:\frac{{dy}}{{dx}}\:=\:\frac{{x}+{y}+{a}}{{x}+{y}+{b}} \\ $$
Answered by mr W last updated on 08/Feb/24
![u=x+y (du/dx)=1+(dy/dx) ((u−a)/(u−b))×((du/dx)−1)=((u+a)/(u+b)) (du/dx)=((u^2 +(a−b)u−ab)/(u^2 −(a−b)u−ab))+1 (du/dx)=((2(u^2 −ab))/(u^2 −(a−b)u−ab)) ((u^2 −(a−b)u−ab)/(u^2 −ab))du=2dx (1−(((a−b)u)/(u^2 −ab)))du=2dx u−(((a−b)ln (u^2 −ab))/2)=2x+C 2u−(a−b)ln (u^2 −ab)=4x+C ⇒2(y−x)−(a−b)ln [(x+y)^2 −ab]=C](https://www.tinkutara.com/question/Q204184.png)
$${u}={x}+{y} \\ $$$$\frac{{du}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}} \\ $$$$\frac{{u}−{a}}{{u}−{b}}×\left(\frac{{du}}{{dx}}−\mathrm{1}\right)=\frac{{u}+{a}}{{u}+{b}} \\ $$$$\frac{{du}}{{dx}}=\frac{{u}^{\mathrm{2}} +\left({a}−{b}\right){u}−{ab}}{{u}^{\mathrm{2}} −\left({a}−{b}\right){u}−{ab}}+\mathrm{1} \\ $$$$\frac{{du}}{{dx}}=\frac{\mathrm{2}\left({u}^{\mathrm{2}} −{ab}\right)}{{u}^{\mathrm{2}} −\left({a}−{b}\right){u}−{ab}} \\ $$$$\frac{{u}^{\mathrm{2}} −\left({a}−{b}\right){u}−{ab}}{{u}^{\mathrm{2}} −{ab}}{du}=\mathrm{2}{dx} \\ $$$$\left(\mathrm{1}−\frac{\left({a}−{b}\right){u}}{{u}^{\mathrm{2}} −{ab}}\right){du}=\mathrm{2}{dx} \\ $$$${u}−\frac{\left({a}−{b}\right)\mathrm{ln}\:\left({u}^{\mathrm{2}} −{ab}\right)}{\mathrm{2}}=\mathrm{2}{x}+{C} \\ $$$$\mathrm{2}{u}−\left({a}−{b}\right)\mathrm{ln}\:\left({u}^{\mathrm{2}} −{ab}\right)=\mathrm{4}{x}+{C} \\ $$$$\Rightarrow\mathrm{2}\left({y}−{x}\right)−\left({a}−{b}\right)\mathrm{ln}\:\left[\left({x}+{y}\right)^{\mathrm{2}} −{ab}\right]={C} \\ $$
Commented by naka3546 last updated on 09/Feb/24
$$ \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much},\:\mathrm{sir}. \\ $$