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Question Number 127260 by bramlexs22 last updated on 28/Dec/20

(x^2 −1)y′′−2xy′+2y = (x^2 −1)^2

$$\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}''−\mathrm{2}{xy}'+\mathrm{2}{y}\:=\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$

Answered by liberty last updated on 28/Dec/20

let y = xz for some z=z(x) y′=xz′+z ∧ y′′=xz′′+2z′ substituting this into the original ODE give (x^2 −1)(xz′′+2z′)−2x(xz′+z)+2xz=(x^2 −1)^2 x(x^2 −1)z′′−2z′=(x^2 −1)^2 z′′−(2/(x(x^2 −1))) z′=((x^2 −1)/x) put integrating factor λ(x)=e^(∫−(2/(x(x^2 −1))) dx) λ(x)=e^(2ln x−ln (x+1)−ln (x−1)) = (x^2 /(x^2 −1)) this gives us (x^2 /(x^2 −1)) z′′−(2/((x^2 −1)^2 )) z′=x (d/dx) ((x^2 /(x^2 −1)) z′) = x ; solving for z′ ⇔(x^2 /(x^2 −1)) z′ = (1/2)x^2 +C_1 or z′ = (1/2)(x^2 −1)+C_1 (x+(1/x)) integrating both sides z = (1/6)(x^3 −3x)+C_1 (x+(1/x))+C_2 (y/x) = (1/6)(x^3 −3x)+C_1 (x+(1/x))+C_2 y = (1/6)(x^4 −3x^2 )+C_1 (x^2 +1)+C_2 x

$${let}\:{y}\:=\:{xz}\:{for}\:{some}\:{z}={z}\left({x}\right) \\ $$$${y}'={xz}'+{z}\:\wedge\:{y}''={xz}''+\mathrm{2}{z}' \\ $$$${substituting}\:{this}\:{into}\:{the}\:{original}\:{ODE} \\ $$$${give}\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left({xz}''+\mathrm{2}{z}'\right)−\mathrm{2}{x}\left({xz}'+{z}\right)+\mathrm{2}{xz}=\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$${x}\left({x}^{\mathrm{2}} −\mathrm{1}\right){z}''−\mathrm{2}{z}'=\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$$${z}''−\frac{\mathrm{2}}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:{z}'=\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}} \\ $$$${put}\:{integrating}\:{factor}\:\lambda\left({x}\right)={e}^{\int−\frac{\mathrm{2}}{{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:{dx}} \\ $$$$\lambda\left({x}\right)={e}^{\mathrm{2ln}\:{x}−\mathrm{ln}\:\left({x}+\mathrm{1}\right)−\mathrm{ln}\:\left({x}−\mathrm{1}\right)} \:=\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${this}\:{gives}\:{us}\:\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:\:{z}''−\frac{\mathrm{2}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:{z}'={x} \\ $$$$\frac{{d}}{{dx}}\:\left(\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:{z}'\right)\:=\:{x}\:;\:{solving}\:{for}\:{z}' \\ $$$$\Leftrightarrow\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{1}}\:{z}'\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +{C}_{\mathrm{1}} \\ $$$${or}\:{z}'\:=\:\frac{\mathrm{1}}{\mathrm{2}}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+{C}_{\mathrm{1}} \left({x}+\frac{\mathrm{1}}{{x}}\right) \\ $$$${integrating}\:{both}\:{sides}\: \\ $$$$\:{z}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)+{C}_{\mathrm{1}} \left({x}+\frac{\mathrm{1}}{{x}}\right)+{C}_{\mathrm{2}} \\ $$$$\:\:\frac{{y}}{{x}}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\left({x}^{\mathrm{3}} −\mathrm{3}{x}\right)+{C}_{\mathrm{1}} \left({x}+\frac{\mathrm{1}}{{x}}\right)+{C}_{\mathrm{2}} \\ $$$$\:\:{y}\:=\:\frac{\mathrm{1}}{\mathrm{6}}\left({x}^{\mathrm{4}} −\mathrm{3}{x}^{\mathrm{2}} \right)+{C}_{\mathrm{1}} \left({x}^{\mathrm{2}} +\mathrm{1}\right)+{C}_{\mathrm{2}} {x}\: \\ $$$$ \\ $$

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