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Question Number 77537 by john santu last updated on 07/Jan/20

what is the solution xy′+(1+xcot x)y=x ?

$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{solution}\: \\ $$$$\mathrm{xy}'+\left(\mathrm{1}+\mathrm{xcot}\:\mathrm{x}\right)\mathrm{y}=\mathrm{x}\:? \\ $$

Answered by mr W last updated on 07/Jan/20

y′+((1/x)+((cos x)/(sin x)))y=1 type y′+p(x)y=q(x) ∫p(x)dx=∫((1/x)+((cos x)/(sin x)))dx=ln (x)+ln (sin x)=ln (x sin x) u(x)=e^(∫p(x)dx) =e^(ln (x sin x)) =x sin x ∫q(x)u(x)dx=∫x sin x dx=∫x d(−cos x)=−x cos x+∫cos x dx=−x cos x+sin x y=((∫q(x)u(x)dx+C)/(u(x)))=((sin x−x cos x +C)/(x sin x))

$${y}'+\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\right){y}=\mathrm{1} \\ $$$${type}\:{y}'+{p}\left({x}\right){y}={q}\left({x}\right) \\ $$$$\int{p}\left({x}\right){dx}=\int\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}}\right){dx}=\mathrm{ln}\:\left({x}\right)+\mathrm{ln}\:\left(\mathrm{sin}\:{x}\right)=\mathrm{ln}\:\left({x}\:\mathrm{sin}\:{x}\right) \\ $$$${u}\left({x}\right)={e}^{\int{p}\left({x}\right){dx}} ={e}^{\mathrm{ln}\:\left({x}\:\mathrm{sin}\:{x}\right)} ={x}\:\mathrm{sin}\:{x} \\ $$$$\int{q}\left({x}\right){u}\left({x}\right){dx}=\int{x}\:\mathrm{sin}\:{x}\:{dx}=\int{x}\:{d}\left(−\mathrm{cos}\:{x}\right)=−{x}\:\mathrm{cos}\:{x}+\int\mathrm{cos}\:{x}\:{dx}=−{x}\:\mathrm{cos}\:{x}+\mathrm{sin}\:{x} \\ $$$${y}=\frac{\int{q}\left({x}\right){u}\left({x}\right){dx}+{C}}{{u}\left({x}\right)}=\frac{\mathrm{sin}\:{x}−{x}\:\mathrm{cos}\:{x}\:+{C}}{{x}\:\mathrm{sin}\:{x}} \\ $$

Commented by mr W last updated on 07/Jan/20

Commented by john santu last updated on 08/Jan/20

thanks you sir

$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$

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