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Consider the linear ODE [itex] y''-(ax-b)y=0 [/itex].
Wolframalpha gives [itex] Ai(\frac{ax-b}{a^{2/3}}) [/itex] and [itex] Bi(\frac{ax-b}{a^{2/3}}) [/itex] as the two independent solutions where Ai(z) and Bi(z) are the Airy functions.
I tried to find an answer myself, so I substituted [itex] y=f(x) e^{cx} [/itex] in the equation and tried to find f. The resulting ODE was hard so I gave it to wolframalpha and upon substituting what I got into y, I got:
[itex]
y=(\frac{ax-b}{c^2}-\frac{2a}{c^3})e^{cx}+k_1 x+k_2
[/itex]
Now I'm confused about the relationship between these solutions. Because a 2nd order ODE has two independent solutions. So the solution I found should somehow be related to Airy functions which I fail to see how. Also it seems to me that there should be another solution independent of the solution I found so that I have two independent solutions too. Then it can be said that I just have two different bases for the space of the ODE's solutions. I doubt it though.
Any ideas?
Thanks
Wolframalpha gives [itex] Ai(\frac{ax-b}{a^{2/3}}) [/itex] and [itex] Bi(\frac{ax-b}{a^{2/3}}) [/itex] as the two independent solutions where Ai(z) and Bi(z) are the Airy functions.
I tried to find an answer myself, so I substituted [itex] y=f(x) e^{cx} [/itex] in the equation and tried to find f. The resulting ODE was hard so I gave it to wolframalpha and upon substituting what I got into y, I got:
[itex]
y=(\frac{ax-b}{c^2}-\frac{2a}{c^3})e^{cx}+k_1 x+k_2
[/itex]
Now I'm confused about the relationship between these solutions. Because a 2nd order ODE has two independent solutions. So the solution I found should somehow be related to Airy functions which I fail to see how. Also it seems to me that there should be another solution independent of the solution I found so that I have two independent solutions too. Then it can be said that I just have two different bases for the space of the ODE's solutions. I doubt it though.
Any ideas?
Thanks