| Linje 1: | Linje 1: | ||
| − | + | 1a) | |
| + | ---- | ||
| + | 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. | ||
| + | |||
| + | 1b) | ||
---- | ---- | ||
| + | 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. | ||
| + | |||
| + | 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>. But y is continuous, and hence there is no place where it can equal 0 (as <tex>Ce^{Rx} \not = 0</tex> for any x, and only tends to zero as Rx tends to negative infinity), so <tex>y=Ce^{Rx}</tex> for all x. It is easily verified that these are all solutions, so they are all the solutions. | ||
Revisjonen fra 3. feb. 2011 kl. 22:52
1a)
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.
1b)
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.
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{yy-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>. But y is continuous, and hence there is no place where it can equal 0 (as <tex>Ce^{Rx} \not = 0</tex> for any x, and only tends to zero as Rx tends to negative infinity), so <tex>y=Ce^{Rx}</tex> for all x. It is easily verified that these are all solutions, so they are all the solutions.
