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Question Number 73495 by ajfour last updated on 13/Nov/19

$$\frac{d^{2}y}{{dx}^2}=c^2{x}^{2}y(y=a,x=0)$$

Answered by mind is power last updated on 13/Nov/19

$$t=x^2$$$$\frac{dy}{dx}=\frac{dy}{dt}.\frac{dt}{dx}=2x\frac{dy}{dt}$$$$\frac{d^{2}y}{{dx}^2}=2\frac{dy}{dt}+2x.\frac{d^{2}y}{{dt}^2}.\frac{dt}{dx}=2\frac{dy}{dt}+4x^2\frac{d^{2}y}{{dt}^2}$$$$\iff 4x^2\frac{d^{2}y}{{dt}^2}+2\frac{dy}{dt}=c^2{x}^{2}y$$$$\iff 4t\frac{d^{2}y}{{dt}^2}+2\frac{dy}{dt}-c^{2}ty\left(t\right)=0$$$$\iff t^2\frac{d^{2}y}{{dt}^2}+\frac{1}{2}t\frac{dy}{dt}-\frac{c^2}{4}{t}^{2}y\left(t\right)=0$$$$BesselgeneraliseEquation$$$$is{x}^2\frac{d^{2}y}{{dx}^2}+(2p+1)x\frac{dy}{dx}+(a^2{x}^{2r}+{\beta }^2)y=0$$$$solutionisy=x^{-p}[C_1{J}_{q/r}\left(\frac{{ax}^r}{r}\right)+C_2{Y}_{q/r}\left(\frac{\alpha }{r}{x}^r\right)]$$$$q=\sqrt{p^2-{\beta }^2}$$$$inourcaser=1$$$$a=\left(\frac{ci}{2}\right)$$$$p=-\frac{1}{4}$$$$q=\sqrt{\frac{1}{16}-0}=\frac{1}{4}$$$$y=t^{\frac{1}{4}}[C_1{J}_{\frac{1}{4}}\left(\frac{ci}{2}t\right)+C_2{Y}_{\frac{1}{4}}\left(\frac{cit}{2}\right)]$$$$Jfirstbasselfunction,Y2ndespaceofbasselequation$$$$y\left(x\right)=Y\left(t\right){\mid }_{t=\sqrt{x}}$$

Commented by ajfour last updated on 13/Nov/19

$$Thanks{}^{\prime }Powerful{Mind}^{\prime },i^{\prime }ll$$$$lookinto{}^{\prime }Erwin{Kreyszig}^{\prime }$$$$(higherengg.mathsbook)and$$$$trytounderstandyour{sol}^n..$$

Commented by mind is power last updated on 13/Nov/19

$$y^{\prime }reWelcomihadapdfofdifferentialequation$$$$ififindititellyoutheNames$$

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