Question Number 27801 by abdo imad last updated on 14/Jan/18
$${solve}\:{the}\:{d}.{e}.\:\:{y}^{'} \:−\mathrm{2}{xy}\:={sinx}\:{e}^{{x}^{\mathrm{2}} } \:{with}\:{initial}\:{condition} \\ $$$${y}\left({o}\right)=\mathrm{1}. \\ $$
Commented by abdo imad last updated on 19/Jan/18
$${he}\:\Rightarrow\:{y}^{'} −\mathrm{2}{xy}=\mathrm{0}\:\Leftrightarrow\:\:\:\frac{{y}^{'} }{{y}}=\mathrm{2}{x}\:\:\Leftrightarrow\:\:{ln}/{y}/\:=\:{x}^{\mathrm{2}} \:+{k} \\ $$$$\Leftrightarrow\:\:{y}=\:\lambda\:{e}^{{x}^{\mathrm{2}} } \:\:\:\:{let}\:{use}\:{the}\:{m}.{v}.{c}\:{method} \\ $$$${y}^{'} =\:\lambda^{'} \:{e}^{{x}^{\mathrm{2}} } \:+\mathrm{2}\lambda{x}\:{e}^{{x}^{\mathrm{2}} } \\ $$$${and}\:\:\left({e}\right)\:\Leftrightarrow\:\:\lambda^{'} \:{e}^{{x}^{\mathrm{2}} } \:+\mathrm{2}\lambda{x}\:{e}^{{x}^{\mathrm{2}} } \:−\mathrm{2}\lambda{x}\:{e}^{{x}^{\mathrm{2}} } =\:{sinx}\:{e}^{{x}^{\mathrm{2}} } \\ $$$$\Leftrightarrow\:\lambda^{'} \:={sinx}\:\:\Leftrightarrow\:\:\:\lambda\:=\:\int{sinxdx}\:+{k}\:=−{cosx}\:+{k}\:{and} \\ $$$${y}\left({x}\right)=\:\left(−{cosx}\:+{k}\right){e}^{{x}^{\mathrm{2}} } \:\:{y}\left(\mathrm{0}\right)=\mathrm{1}\:\Rightarrow{k}−\mathrm{1}=\mathrm{1}\:\Rightarrow{k}=\mathrm{2} \\ $$$${finally}\:\:\:{y}\left({x}\right)=\:\left(\mathrm{2}−{cosx}\right){e}^{{x}^{\mathrm{2}} } \:\:. \\ $$