Question Number 97137 by mhmd last updated on 06/Jun/20
$${using}\:{aparticular}\:{theory}\:,{find}\:{the}\:{general}\:{solution}\:{to}\: \\ $$$${the}\:{following}\:{differential}\:{equation}\: \\ $$$${f}\left({x}+{y}\right){dx}+{g}\left({x}+{y}\right){dy}=\mathrm{0}\:? \\ $$$${help}\:{me}\:{sir}\:{please} \\ $$
Commented by prakash jain last updated on 06/Jun/20
$$\mathrm{what}\:\mathrm{do}\:\mathrm{u}\:\mathrm{mean}\:\mathrm{by}\:\mathrm{a}\:\mathrm{particular} \\ $$$$\mathrm{theory}? \\ $$
Commented by mhmd last updated on 07/Jun/20
$${Specific}\:{theorem}\:{sir}\: \\ $$
Answered by mr W last updated on 07/Jun/20
$$\frac{{dy}}{{dx}}=−\frac{{f}\left({x}+{y}\right)}{{g}\left({x}+{y}\right)} \\ $$$${let}\:{u}={x}+{y} \\ $$$$\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}−\mathrm{1} \\ $$$$\frac{{du}}{{dx}}−\mathrm{1}=−\frac{{f}\left({u}\right)}{{g}\left({u}\right)} \\ $$$$\frac{{g}\left({u}\right){du}}{{g}\left({u}\right)−{f}\left({u}\right)}={dx} \\ $$$$\Rightarrow{x}=\int\frac{{g}\left({u}\right){du}}{{g}\left({u}\right)−{f}\left({u}\right)}={h}\left({u}\right)+{C} \\ $$$$\Rightarrow{x}={h}\left({x}+{y}\right)+{C} \\ $$