Question Number 118349 by syamil last updated on 17/Oct/20
$${solution}\:{xdy}\:+\:\left(\mathrm{3}{y}\:−\:{e}^{{x}} \right){dx}\: \\ $$$${with}\:{integration}\:{factor} \\ $$
Answered by Dwaipayan Shikari last updated on 17/Oct/20
$${xdy}+\left(\mathrm{3}{y}−{e}^{{x}} \right){dx}=\mathrm{0} \\ $$$${x}\frac{{dy}}{{dx}}+\mathrm{3}{y}−{e}^{{x}} =\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}+\frac{\mathrm{3}{y}}{{x}}={e}^{{x}} \\ $$$${I}.{F}={e}^{\int\frac{\mathrm{3}}{{x}}} ={x}^{\mathrm{3}} \\ $$$${y}.{x}^{\mathrm{3}} =\int{x}^{\mathrm{3}} {e}^{{x}} \\ $$$${y}.{x}^{\mathrm{3}} ={x}^{\mathrm{3}} {e}^{{x}} −\int\mathrm{3}{x}^{\mathrm{2}} {e}^{{x}} \\ $$$${y}.{x}^{\mathrm{3}} ={x}^{\mathrm{3}} {e}^{{x}} −\mathrm{3}{x}^{\mathrm{2}} {e}^{{x}} +\mathrm{6}\int{xe}^{{x}} \\ $$$${y}.{x}^{\mathrm{3}} ={x}^{\mathrm{3}} {e}^{{x}} −\mathrm{3}{x}^{\mathrm{2}} {e}^{{x}} +\mathrm{6}{xe}^{{x}} −\mathrm{6}{e}^{{x}} +{C} \\ $$$${y}={e}^{{x}} \left(\mathrm{1}−\frac{\mathrm{3}}{{x}}+\frac{\mathrm{6}}{{x}^{\mathrm{2}} }−\frac{\mathrm{6}}{{x}^{\mathrm{3}} }\right)+{Cx}^{−\mathrm{3}} \\ $$$$ \\ $$