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Question Number 85165 by mathmax by abdo last updated on 19/Mar/20
$${sove}\:\:\left({sin}^{\mathrm{2}} {x}\right)\:{y}^{'} \:\:+\left({cosx}\right){y}\:={x} \\ $$
Commented by mathmax by abdo last updated on 20/Mar/20
$$\left({he}\right)\rightarrow\left({sin}^{\mathrm{2}} {x}\right){y}^{'} \:+\left({cosx}\right){y}\:=\mathrm{0}\:\Rightarrow\frac{{y}^{'} }{{y}}\:=−\frac{{cosx}}{{sin}^{\mathrm{2}} {x}}\:\Rightarrow \\ $$$${ln}\mid{y}\mid\:=\frac{\mathrm{1}}{{sinx}}\:+{k}\:\Rightarrow{y}\left({x}\right)={C}\:{e}^{\frac{\mathrm{1}}{{sinx}}} \:\:\:{let}\:{use}\:{mvc}\:{method} \\ $$$${y}^{'} \left({x}\right)={C}^{'} \:{e}^{\frac{\mathrm{1}}{{sinx}}} \:+{C}\left(−\frac{{cosx}}{{sin}^{\mathrm{2}} {x}}\right){e}^{\frac{\mathrm{1}}{{sinx}}} \\ $$$$\left({e}\right)\:\Rightarrow{C}^{'} \:{sin}^{\mathrm{2}} {x}\:{e}^{\frac{\mathrm{1}}{{sinx}}} \:−{C}\:{cosx}\:{e}^{\frac{\mathrm{1}}{{sinx}}} \:+{C}\:{cosx}\:{e}^{\frac{\mathrm{1}}{{sinx}}} \:={x}\:\Rightarrow \\ $$$${C}\:^{'} \:=\frac{{x}}{{sin}^{\mathrm{2}} {x}}{e}^{−\frac{\mathrm{1}}{{sinx}}} \:\Rightarrow{C}\left({x}\right)\:=\:\int^{{x}} \frac{{u}}{{sin}^{\mathrm{2}} {u}}\:{e}^{−\frac{\mathrm{1}}{{sinu}}} \:{du}\:\:+\lambda\:\Rightarrow \\ $$$${y}\left({x}\right)\:=\left(\int^{{x}} \:\frac{{u}}{{sin}^{\mathrm{2}} {u}}{e}^{−\frac{\mathrm{1}}{{sinu}}} {du}\:+\lambda\right){e}^{\frac{\mathrm{1}}{{sinx}}} \:{is}\:{the}\:{general}\left[{solution}\right. \\ $$