Question Number 42221 by rahul 19 last updated on 20/Aug/18
$$\mathrm{Solve}: \\ $$$$\frac{\mathrm{dt}}{\mathrm{d}{x}}\:=\:\frac{\mathrm{2}}{{x}+\mathrm{t}}\:. \\ $$
Answered by MrW3 last updated on 20/Aug/18
$${u}={x}+{t} \\ $$$$\frac{{du}}{{dx}}=\mathrm{1}+\frac{{dt}}{{dx}} \\ $$$$\Rightarrow\frac{{du}}{{dt}}−\mathrm{1}=\frac{\mathrm{2}}{{u}} \\ $$$$\Rightarrow\frac{{du}}{{dt}}=\frac{\mathrm{2}+{u}}{{u}} \\ $$$$\Rightarrow\frac{{udu}}{\mathrm{2}+{u}}={dt} \\ $$$$\Rightarrow\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{2}+{u}}\right){du}={dt} \\ $$$$\Rightarrow{u}−\mathrm{2}\:\mathrm{ln}\:\left(\mathrm{2}+{u}\right)={t}+{C} \\ $$$$\Rightarrow\left({x}+{t}\right)−\mathrm{2}\:\mathrm{ln}\:\left(\mathrm{2}+{x}+{t}\right)={t}+{C} \\ $$$$\Rightarrow{x}=\mathrm{2}\:\mathrm{ln}\:\left(\mathrm{2}+{x}+{t}\right)+{C} \\ $$
Commented by rahul 19 last updated on 20/Aug/18
thanks sir ��