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Linje 1: |
| − | 1a)
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| − | For every x, there exists a d(x) such that <tex>y^{(n)}(x)-y^{(n-1)}(x)=d(x)</tex> by the assumptions of the problem. As y is smooth, so are its derivatives, so d(x) is a difference between two smooth functions and hence smooth. We will now write d=d(x) and y=y(x) for simplicity. Note that d'=<tex>y'-y)'=y''-y'=d</tex>, so d'=d, which can be easily solved for d to obtain <tex>d=Ce^x</tex>. Hence <tex>y'-y=Ce^x</tex>, which is a simple differential equation which yields <tex>y=(Cx+D)e^x</tex>. By insertion we see that these are solutions for all choices of C, D, so these are all the solutions.
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| − | 1b)
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| − | As in the previous part, we see that there exists a function r(x)=r (Again we are not saying that r(x) is constant, but simplifying notation.) such that <tex>y^{(n+1)}=ry^{(n)}</tex>. We see then that r is smooth except possibly where y=0. Note also that in this case <tex>y^{(m)}=0</tex> for all m.
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| − | We work now in <tex>Y=\mathbb {R}</tex> <tex>-y^{-1}(\{0\})</tex>. As <tex>\{0\}</tex> is closed, this is an open set, so we have that r is smooth on this open set, and we can solve the differential equation here, as <tex>y\not = 0</tex>. Then <tex>r'=\left ( \frac {y'} {y} \right ) ' = \frac{y''y-y' \cdot y'} {y^2} = \frac {r^2y^2- (ry)^2} {y^2} = 0</tex>, so r must be constant, and hence <tex>y'=Ry</tex> for some constant R, which is easily solved to yield <tex>y=Ce^{Rx}</tex> for all <tex>x \in Y</tex>.
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| − | Now, assuming y is not constantly equal to 0, Y is nonempty, so select some p in Y. Let <tex>A=(-\infty, p] \cap y^{-1} (\{0 \})</tex> and <tex>B=[p, \infty) \cap y^{-1} (\{0 \})</tex>. Either y is never equal to zero, or one of these are nonempty, so either sup A or inf B exists. Without loss of generality, we suppose sup A exists. Then this is a limit point of Y, so we can find a sequence <tex>c_n \in Y</tex> converging to <tex>\sup A</tex>. But it is also a limit point of A, so we can find a sequence <tex>a_n \in A</tex> converging to <tex>\sup A</tex>. However the former implies that <tex>y(\sup A)=\lim y(y_n) \not = 0</tex> (as <tex>y(y_n)=Ce^{Ry_n}</tex>, which (as C is assumed to be nonzero) only tends to zero if <tex>Ry_n</tex> tends to negative infinity, which is not the case as <tex>y_n</tex> has a real (finite) limit), and the latter implies that <tex>y(\sup A)=\lim y(a_n)=0</tex>, so we have a contradiction. Hence y is either always equal to zero, or never equal to zero, which implies that it is one of the solutions we have found.
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| − | Hence <tex>y=Ce^{Rx}</tex> are all the solutions. By insertion all choices of C, R yield valid solutions.
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| − | Huskepå
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| − | Nå kom det en revisjon nettopp der de sa at den geometriske rekka måtte være ikke-triviell, det vil si med alle ledd forskjellige fra null. I så fall blir løsningen mye enklere i og med at vi kan se bortifra divisjon på null og alt det der, så vi kan egentlig kutte ned på et avsnitt eller to. Vi må også huske på å nevne at C, R skal være forskjellige fra null.
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