Question Number 41378 by rahul 19 last updated on 06/Aug/18
$$\mathrm{Solve}\:: \\ $$$$\mathrm{e}^{{x}} \left({x}+\mathrm{1}\right){dx}\:+\:\left(\mathrm{ye}^{\mathrm{y}} \:−\:{xe}^{{x}} \right)\mathrm{dy}=\mathrm{0} \\ $$
Commented by rahul 19 last updated on 06/Aug/18
If I put xe^x = t. does this helps?
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18
$${u}={e}^{{x}} {x}\:\:{du}={e}^{{x}} \left({x}+\mathrm{1}\right){dx} \\ $$$${du}+\left({ye}^{{y}} −{u}\right){dy}=\mathrm{0} \\ $$$${du}=\left({u}−{ye}^{{y}} \right){dy} \\ $$$$\frac{{du}}{{dy}}−{u}=−{ye}^{{y}} \\ $$$${intregating}\:{factor}\:{e}^{\int−{dy}} ={e}^{−{y}} \\ $$$${e}^{−{y}} \frac{{du}}{{dy}}+\left(−{ue}^{−{y}} \right)=−{y} \\ $$$$\frac{{d}\left({u}.{e}^{−{y}} \right)}{{dy}}=−{y} \\ $$$${d}\left({u}.{e}^{−{y}} \right)=−{ydy} \\ $$$${intregating} \\ $$$${u}.{e}^{−{y}} =−\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$$\left({e}^{{x}} .{x}\right).{e}^{−{y}} =−\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{c} \\ $$$$ \\ $$
Commented by rahul 19 last updated on 07/Aug/18
Thank You sir ������
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18
$${checking}… \\ $$$${differentiating}\:{wrt}\:{to}\:{x} \\ $$$${e}^{−{y}} \left\{{e}^{{x}} \left({x}+\mathrm{1}\right)\right\}+{e}^{{x}} .{x}\left(−{e}^{−{y}} \right)\frac{{dy}}{{dx}}=−{y}\frac{{dy}}{{dx}} \\ $$$${multiply}\:{both}\:{side}\:{by}\:{e}^{{y}} \\ $$$${e}^{{x}} \left({x}+\mathrm{1}\right)−{e}^{{x}} {x}\frac{{dy}}{{dx}}=−{ye}^{{y}} \frac{{dy}}{{dx}} \\ $$$${e}^{{x}} \left({x}+\mathrm{1}\right){dx}+\left({e}^{{y}} {y}−{e}^{{x}} {x}\right){dy}=\mathrm{0} \\ $$
Commented by rahul 19 last updated on 07/Aug/18
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