- #1

yungman

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Problem to solve: [tex]x^{2}y''+xy'+(\lambda^{2}x^{2}-\frac{1}{4})=0[/tex] subject to [tex]y(0)=finite,y(\pi)=0[/tex]

Let [tex]t=\lambda x[/tex] which give [tex]t^{2}\frac{d^{2}y}{dt^{2}}+t\frac{dy}{dt}+(t^{2}-\frac{1}{2})y=0[/tex]

With that first linear independent solution is [tex]J_{\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex] and second independent solution is [tex]J_{-\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex]

Since these two are linear independent solution, I should be able to write the general solution of y:

[tex]y=C_{1}J_{\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)] + C_{2}J_{-\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex]

But the book claimed the solution of y:[tex]y=C_{1}J_{\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)] + C_{2}Y_{-\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex]

Question is why do I have to use Y in the second solution instead of both J to solve the boundary value problem? I always learn that if you get the two independent solution, that would be the answers!

Let [tex]t=\lambda x[/tex] which give [tex]t^{2}\frac{d^{2}y}{dt^{2}}+t\frac{dy}{dt}+(t^{2}-\frac{1}{2})y=0[/tex]

With that first linear independent solution is [tex]J_{\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex] and second independent solution is [tex]J_{-\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex]

Since these two are linear independent solution, I should be able to write the general solution of y:

[tex]y=C_{1}J_{\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)] + C_{2}J_{-\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex]

But the book claimed the solution of y:[tex]y=C_{1}J_{\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)] + C_{2}Y_{-\frac{1}{2}}[\sqrt{\frac{2}{\pi\lambda x}}sin(\lambda x)][/tex]

Question is why do I have to use Y in the second solution instead of both J to solve the boundary value problem? I always learn that if you get the two independent solution, that would be the answers!

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