anonymous
  • anonymous
prove if the set {e^ax, xe^ax} is linearly independent
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
KingGeorge
  • KingGeorge
To show this is linearly independent, you need to let \[a_1e^{ax}+a_2xe^{ax}=0\]and show that \(a_1\) and \(a_2\) are 0.
KingGeorge
  • KingGeorge
In other words, show that \(a_1e^{ax}\neq a_2 xe^{ax}\) for all \(x\). To begin, you might want to multiply by \(e^{-ax}\).
anonymous
  • anonymous
You could also use the Wronskian determinant: http://en.wikipedia.org/wiki/Wronskian

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
I highly recommend using the Wronskian.
anonymous
  • anonymous
And Abel's theorem allows you to simplify it even further because if it vanishes at a point t_0 then it is identically zero for all t.
anonymous
  • anonymous
If I remember DiffEQ correctly haha

Looking for something else?

Not the answer you are looking for? Search for more explanations.