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Question Number 70232 by Joel122 last updated on 02/Oct/19

Solve  xy′ − y sin x + y^5  = 0

$$\mathrm{Solve} \\ $$$${xy}'\:−\:{y}\:\mathrm{sin}\:{x}\:+\:{y}^{\mathrm{5}} \:=\:\mathrm{0} \\ $$

Commented by Joel122 last updated on 02/Oct/19

I think it is a Bernoulli′s equation  y′ − (((sin x)/x))y = −(y^5 /x)  using u = y^(−4)  gives  u′ + (((4sin x)/x))u = (4/x)  and integrating factor F(x) = e^(∫ ((4 sin x)/x) dx)

$$\mathrm{I}\:\mathrm{think}\:\mathrm{it}\:\mathrm{is}\:\mathrm{a}\:\mathrm{Bernoulli}'\mathrm{s}\:\mathrm{equation} \\ $$$${y}'\:−\:\left(\frac{\mathrm{sin}\:{x}}{{x}}\right){y}\:=\:−\frac{{y}^{\mathrm{5}} }{{x}} \\ $$$$\mathrm{using}\:{u}\:=\:{y}^{−\mathrm{4}} \:\mathrm{gives} \\ $$$${u}'\:+\:\left(\frac{\mathrm{4sin}\:{x}}{{x}}\right){u}\:=\:\frac{\mathrm{4}}{{x}} \\ $$$$\mathrm{and}\:\mathrm{integrating}\:\mathrm{factor}\:{F}\left({x}\right)\:=\:{e}^{\int\:\frac{\mathrm{4}\:\mathrm{sin}\:{x}}{{x}}\:{dx}} \\ $$

Commented by Joel122 last updated on 02/Oct/19

Now I′m stuck. Can anyone help? Or  is it any other way to solve this?

$$\mathrm{Now}\:\mathrm{I}'\mathrm{m}\:\mathrm{stuck}.\:\mathrm{Can}\:\mathrm{anyone}\:\mathrm{help}?\:\mathrm{Or} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{any}\:\mathrm{other}\:\mathrm{way}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}? \\ $$

Commented by mind is power last updated on 02/Oct/19

∫((sin(x))/x)=Si(x) Sinus integral function  But after that we cant find particular solution  particular solution will bee by power Series

$$\int\frac{{sin}\left({x}\right)}{{x}}={Si}\left({x}\right)\:{Sinus}\:{integral}\:{function} \\ $$$${But}\:{after}\:{that}\:{we}\:{cant}\:{find}\:{particular}\:{solution} \\ $$$${particular}\:{solution}\:{will}\:{bee}\:{by}\:{power}\:{Series} \\ $$

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