Question Number 77537 by john santu last updated on 07/Jan/20
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}\: \\ $$$$\mathrm{xy}'+\left(\mathrm{1}+\mathrm{xcot}\:\mathrm{x}\right)\mathrm{y}=\mathrm{x}\:? \\ $$
Answered by mr W last updated on 07/Jan/20
$${y}'+\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\right){y}=\mathrm{1} \\ $$$${type}\:{y}'+{p}\left({x}\right){y}={q}\left({x}\right) \\ $$$$\int{p}\left({x}\right){dx}=\int\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\right){dx}=\mathrm{ln}\:\left({x}\right)+\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)=\mathrm{ln}\:\left({x}\:\mathrm{sin}\:{x}\right) \\ $$$${u}\left({x}\right)={e}^{\int{p}\left({x}\right){dx}} ={e}^{\mathrm{ln}\:\left({x}\:\mathrm{sin}\:{x}\right)} ={x}\:\mathrm{sin}\:{x} \\ $$$$\int{q}\left({x}\right){u}\left({x}\right){dx}=\int{x}\:\mathrm{sin}\:{x}\:{dx}=\int{x}\:{d}\left(−\mathrm{cos}\:{x}\right)=−{x}\:\mathrm{cos}\:{x}+\int\mathrm{cos}\:{x}\:{dx}=−{x}\:\mathrm{cos}\:{x}+\mathrm{sin}\:{x} \\ $$$${y}=\frac{\int{q}\left({x}\right){u}\left({x}\right){dx}+{C}}{{u}\left({x}\right)}=\frac{\mathrm{sin}\:{x}−{x}\:\mathrm{cos}\:{x}\:+{C}}{{x}\:\mathrm{sin}\:{x}} \\ $$
Commented by mr W last updated on 07/Jan/20
Commented by john santu last updated on 08/Jan/20
$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$