Vector spaces problem -linear algebra

  • Thread starter manuel325
  • Start date
  • #1
16
0

Homework Statement


Hi guys , I have this problem ,well actually I don't understand the solution they provide , Here's the problem statement and the solution .
linear6.JPG

May someone please explain the solution to me?? Thanks so much, Sorry for my bad english


Homework Equations


1.I understand that f'-af=0 and the kernel is the space of the solutions that satisfy that equation but I don't get what they do after that...why do they divide f(t) by e^(at)?
2. why do they conclude that exists a constant c such that f(t)=ce^at??
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
966
The problem is this: we have the vector space of all infinitely differentiable functions, D is the differentiation operator.
a) Find the kernel of the linear operator D- I.
b) Find the kernel of the linear operator D- aI.

It is simplest to solve (b) first, then take a= 1 to solve (a).
If f is any function in the kernel of D- aI then, by definition of "kernel" we must have f'- af= 0.
That is the same as df/dx= af which is a separable equation: [tex]df/f= adx[/tex]. Integrating both sides, ln(f)= ax+ d where c is the constant of integration. Taking the exponential of both sides [itex]f(x)= e^{ax+ c}= Ce^{ax}[/itex] where [itex]C= e^c[/itex].

That is how I would have solved the problem. I suspect that your text, knowing that f must be an exponential, started from that:
[tex]\left(\frac{f(x)}{e^{ax}}\right)'= \frac{f'(x)e^{ax}- f(x)ae^{ax}}{e^{2ax}}= \frac{(f'(x)- af(x))e^{ax}}{e^{2ax}}[/tex]
by the product rule.

And, because [itex]f'(x)- af(x)= 0[/itex], the right side is 0, the derivative of [itex]f(x)/e^{ax}[/itex] is 0 so that [itex]f(x)/e^{ax}[/itex] is a constant: [itex]f(x)/e^{ax}= C[/itex] so [itex]f(x)= Ce^{ax}[/itex].

(And your English is excellent. Far better than my (put language of your choice here).
 
  • #3
16
0
Thank you so much HallsofIvy , I understood all your explanation . It's cool to know that there's still good people who like to help others .Greetings from Chile
 

Related Threads on Vector spaces problem -linear algebra

  • Last Post
Replies
5
Views
5K
  • Last Post
Replies
20
Views
3K
  • Last Post
2
Replies
26
Views
3K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
6
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
3
Views
1K
Replies
6
Views
3K
  • Last Post
Replies
0
Views
1K
Top