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Question Number 97137 by mhmd last updated on 06/Jun/20

using aparticular theory ,find the general solution to   the following differential equation   f(x+y)dx+g(x+y)dy=0 ?  help me sir please

$${using}\:{aparticular}\:{theory}\:,{find}\:{the}\:{general}\:{solution}\:{to}\: \\ $$$${the}\:{following}\:{differential}\:{equation}\: \\ $$$${f}\left({x}+{y}\right){dx}+{g}\left({x}+{y}\right){dy}=\mathrm{0}\:? \\ $$$${help}\:{me}\:{sir}\:{please} \\ $$

Commented by prakash jain last updated on 06/Jun/20

what do u mean by a particular  theory?

$$\mathrm{what}\:\mathrm{do}\:\mathrm{u}\:\mathrm{mean}\:\mathrm{by}\:\mathrm{a}\:\mathrm{particular} \\ $$$$\mathrm{theory}? \\ $$

Commented by mhmd last updated on 07/Jun/20

Specific theorem sir

$${Specific}\:{theorem}\:{sir}\: \\ $$

Answered by mr W last updated on 07/Jun/20

(dy/dx)=−((f(x+y))/(g(x+y)))  let u=x+y  (dy/dx)=(du/dx)−1  (du/dx)−1=−((f(u))/(g(u)))  ((g(u)du)/(g(u)−f(u)))=dx  ⇒x=∫((g(u)du)/(g(u)−f(u)))=h(u)+C  ⇒x=h(x+y)+C

$$\frac{{dy}}{{dx}}=−\frac{{f}\left({x}+{y}\right)}{{g}\left({x}+{y}\right)} \\ $$$${let}\:{u}={x}+{y} \\ $$$$\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}−\mathrm{1} \\ $$$$\frac{{du}}{{dx}}−\mathrm{1}=−\frac{{f}\left({u}\right)}{{g}\left({u}\right)} \\ $$$$\frac{{g}\left({u}\right){du}}{{g}\left({u}\right)−{f}\left({u}\right)}={dx} \\ $$$$\Rightarrow{x}=\int\frac{{g}\left({u}\right){du}}{{g}\left({u}\right)−{f}\left({u}\right)}={h}\left({u}\right)+{C} \\ $$$$\Rightarrow{x}={h}\left({x}+{y}\right)+{C} \\ $$

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