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Wronskian question

  1. Aug 12, 2008 #1
    "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.
     
  2. jcsd
  3. Aug 12, 2008 #2

    Dick

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    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?
     
  4. Aug 12, 2008 #3
    is that supposed to say W(x,f(g),f(h)) on the left side?
     
  5. Aug 12, 2008 #4

    Dick

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