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Question Number 124923 by bemath last updated on 07/Dec/20

 (dy/dx) −y = 3cot x .e^(sin x)

$$\:\frac{{dy}}{{dx}}\:−{y}\:=\:\mathrm{3cot}\:{x}\:.{e}^{\mathrm{sin}\:{x}} \: \\ $$$$ \\ $$

Commented by mohammad17 last updated on 07/Dec/20

P(x)=−1     ,    Q(x)=3cotx.e^(sinx)     (I.f)=e^(∫P(x)dx) =e^(∫−dx) =e^(−x)     ye^(−x) =∫(I.f)Q(x)dx    ye^(−x) =3∫cotx.e^(sinx−1) dx    ye^(−x) =(1/e)Ei(sinx)+C    y=e^(x−1) Ei(sinx)+Ce^x     (Mohammad Al−dolaimy)

$${P}\left({x}\right)=−\mathrm{1}\:\:\:\:\:,\:\:\:\:{Q}\left({x}\right)=\mathrm{3}{cotx}.{e}^{{sinx}} \\ $$$$ \\ $$$$\left({I}.{f}\right)={e}^{\int{P}\left({x}\right){dx}} ={e}^{\int−{dx}} ={e}^{−{x}} \\ $$$$ \\ $$$${ye}^{−{x}} =\int\left({I}.{f}\right){Q}\left({x}\right){dx} \\ $$$$ \\ $$$${ye}^{−{x}} =\mathrm{3}\int{cotx}.{e}^{{sinx}−\mathrm{1}} {dx} \\ $$$$ \\ $$$${ye}^{−{x}} =\frac{\mathrm{1}}{{e}}{Ei}\left({sinx}\right)+{C} \\ $$$$ \\ $$$${y}={e}^{{x}−\mathrm{1}} {Ei}\left({sinx}\right)+{Ce}^{{x}} \\ $$$$ \\ $$$$\left({Mohammad}\:{Al}−{dolaimy}\right) \\ $$

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