- #1
AdkinsJr
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Edit: I think I may have posted this in the wrong section, sorry about that. Note that this isn't a homework problem though, I"m not enrolled in this class, I was just reading over some of this stuff and trying some problems since I"m majoring in physics.
I have a textbook "discussion" problem that's stumping me, I'm given that these are two linearly independent solutions:
[tex]y_1=x^3[/tex] and [tex]y_2=\mid x\mid ^3[/tex]
to the homogeneous differential equation [tex]x^2y''-4xy'+6y=0[/tex]
on [tex](-\infty,\infty)[/tex]
I'm asked to show that the wronskian matrix for the solutions is equal to zero for every real number x. This is easy enough to do, I won't show that work here... but then I"m asked whether this violates the theorem for the wronskian matrix test, which is stated in my text:
If [tex]y_1,y_2...y_n[/tex] are n solutions of a homogenous linear nth-order differential equation on an interval I. Then the set of solutions is linearly independent on I if and if [tex]w(y_1,y_2...y_n)≠0[/tex] for every x in the interval. Naturally I would say yes, it is a violation, but they seem to be implying that it isn't a violation (i can just tell by the way they're asking...). They're messing with me. Why is it not a violation?
I have a textbook "discussion" problem that's stumping me, I'm given that these are two linearly independent solutions:
[tex]y_1=x^3[/tex] and [tex]y_2=\mid x\mid ^3[/tex]
to the homogeneous differential equation [tex]x^2y''-4xy'+6y=0[/tex]
on [tex](-\infty,\infty)[/tex]
I'm asked to show that the wronskian matrix for the solutions is equal to zero for every real number x. This is easy enough to do, I won't show that work here... but then I"m asked whether this violates the theorem for the wronskian matrix test, which is stated in my text:
If [tex]y_1,y_2...y_n[/tex] are n solutions of a homogenous linear nth-order differential equation on an interval I. Then the set of solutions is linearly independent on I if and if [tex]w(y_1,y_2...y_n)≠0[/tex] for every x in the interval. Naturally I would say yes, it is a violation, but they seem to be implying that it isn't a violation (i can just tell by the way they're asking...). They're messing with me. Why is it not a violation?