Question Number 227272 by Spillover last updated on 11/Jan/26
Answered by breniam last updated on 12/Jan/26
$$\left({x}+\mathrm{1}\right){y}'\left({x}\right)−{y}\left({x}\right)={e}^{{x}} \left(\mathrm{1}+{x}\right)^{\mathrm{2}} \\ $$$${y}\left({x}\right)={a}\left({x}\right){b}\left({x}\right) \\ $$$${y}'\left({x}\right)={a}'\left({x}\right){b}\left({x}\right)+{a}\left({x}\right){b}'\left({x}\right) \\ $$$${a}\left({x}\right)\left(\left({x}+\mathrm{1}\right){b}'\left({x}\right)−{b}\left({x}\right)\right)+{a}'\left({x}\right){b}\left({x}\right)\left({x}+\mathrm{1}\right) \\ $$$$\left({x}+\mathrm{1}\right){b}'\left({x}\right)−{b}\left({x}\right)=\mathrm{0} \\ $$$$\frac{{b}'\left({x}\right)}{{b}\left({x}\right)}=\frac{\mathrm{1}}{{x}+\mathrm{1}} \\ $$$$\int\frac{{b}'\left({x}\right)}{{b}\left({x}\right)}\mathrm{d}{x}=\left\{{t}={b}\left({x}\right)\right\}=\int\frac{\mathrm{1}}{{t}}\mathrm{d}{t}=\mathrm{ln}\mid{t}\mid=\mathrm{ln}\mid{b}\left({x}\right)\mid=\int\frac{\mathrm{1}}{{x}+\mathrm{1}}\mathrm{d}{x}=\mathrm{ln}\mid{x}+\mathrm{1}\mid\Rightarrow{b}\left({x}\right)={x}+\mathrm{1} \\ $$$${a}'\left({x}\right)\left({x}+\mathrm{1}\right)={e}^{{x}} \left(\mathrm{1}+{x}\right)^{\mathrm{2}} \\ $$$${a}'\left({x}\right)={e}^{{x}} \left(\mathrm{1}+{x}\right) \\ $$$${a}\left({x}\right)=\int{e}^{{x}} \left(\mathrm{1}+{x}\right)\mathrm{d}{x}={e}^{{x}} \left(\mathrm{1}+{x}\right)−{e}^{{x}} +{A}={xe}^{{x}} +{A} \\ $$$${y}\left({x}\right)=\left({x}+\mathrm{1}\right)\left({xe}^{{x}} +{A}\right) \\ $$