Wronskian question

  • Thread starter jaejoon89
  • Start date
  • #1
195
0
"For the Wronskian, W, Show W(x,fg,fh)=([f(x)]^2)W(g,h)"

How is this done? I know how to use the Wronskian when there's a system of equations, something like y(x) = cosx, y(x)=sinx, y(x)=x, etc. But I'm really clueless about how to proceed here.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,263
619
W(x,fg,fh)=det([[fg,fh],[(fg)',(fh)']]). Right? Now use properties of the determinant. You can factor f out of the first row. Now you have f*det([[g,h],[(fg)',(fh)']]). Use the product rule on the second row. Now you can multiply any row of a determinant by a factor and add it to another row without changing the determinant. Also right? Add -f' times the first row to the second. Getting it yet?
 
  • #3
75
0
is that supposed to say W(x,f(g),f(h)) on the left side?
 
  • #4
Dick
Science Advisor
Homework Helper
26,263
619
is that supposed to say W(x,f(g),f(h)) on the left side?

No. It's W(x,f(x)*g(x),f(x)*h(x)). Otherwise it doesn't work.
 

Related Threads on Wronskian question

  • Last Post
Replies
5
Views
3K
  • Last Post
Replies
2
Views
980
Replies
7
Views
3K
Replies
2
Views
2K
  • Last Post
Replies
2
Views
920
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
23
Views
2K
  • Last Post
Replies
2
Views
975
Top