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Question Number 90641 by Tony Lin last updated on 25/Apr/20

Solve x^2 y′′+xy′+x^2 y=0

$${Solve}\:{x}^{\mathrm{2}} {y}''+{xy}'+{x}^{\mathrm{2}} {y}=\mathrm{0} \\ $$

Commented by jagoll last updated on 25/Apr/20

Euler − Cauchy?

$${Euler}\:−\:{Cauchy}? \\ $$

Commented by Joel578 last updated on 25/Apr/20

nope. It is Bessel differential equation  x^2  (d^2 y/dx^2 ) + x (dy/dx) + (x^2  − ν^2 )y(x) = 0  General solution if ν = non integer  y(x) = c_1 J_ν (x) + c_2 J_(−ν) (x)  If ν = integer  y(x) = c_1 J_ν (x) + c_2 Y_ν (x)    J_ν (x) : Σ_(m=0) ^∞  (((−1)^m )/(m! (m+ν)!)) ((x/2))^(2m+ν)   Y_ν (x) : ((cos (νπ)J_ν (x) − J_(−ν) (x))/(sin (νπ)))

$${nope}.\:\mathrm{It}\:\mathrm{is}\:\mathrm{Bessel}\:\mathrm{differential}\:\mathrm{equation} \\ $$$${x}^{\mathrm{2}} \:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\:+\:{x}\:\frac{{dy}}{{dx}}\:+\:\left({x}^{\mathrm{2}} \:−\:\nu^{\mathrm{2}} \right){y}\left({x}\right)\:=\:\mathrm{0} \\ $$$$\mathrm{General}\:\mathrm{solution}\:\mathrm{if}\:\nu\:=\:\mathrm{non}\:\mathrm{integer} \\ $$$${y}\left({x}\right)\:=\:{c}_{\mathrm{1}} {J}_{\nu} \left({x}\right)\:+\:{c}_{\mathrm{2}} {J}_{−\nu} \left({x}\right) \\ $$$$\mathrm{If}\:\nu\:=\:\mathrm{integer} \\ $$$${y}\left({x}\right)\:=\:{c}_{\mathrm{1}} {J}_{\nu} \left({x}\right)\:+\:{c}_{\mathrm{2}} {Y}_{\nu} \left({x}\right) \\ $$$$ \\ $$$${J}_{\nu} \left({x}\right)\::\:\underset{{m}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{{m}} }{{m}!\:\left({m}+\nu\right)!}\:\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}{m}+\nu} \\ $$$${Y}_{\nu} \left({x}\right)\::\:\frac{\mathrm{cos}\:\left(\nu\pi\right){J}_{\nu} \left({x}\right)\:−\:{J}_{−\nu} \left({x}\right)}{\mathrm{sin}\:\left(\nu\pi\right)} \\ $$

Commented by Tony Lin last updated on 25/Apr/20

thanks sir

$${thanks}\:{sir} \\ $$

Commented by Joel578 last updated on 25/Apr/20

sorry. edit small typo

$${sorry}.\:{edit}\:{small}\:{typo} \\ $$

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