The Wronskian and diff.equations

  • Thread starter Susanne217
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  • #1
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Hi

Homework Statement



I have differential equation

[tex]y'' + p(x)\cdot x' + q(x)y = 0[/tex] which have two solutions [tex]y_1(x)[/tex] and [tex]y_2(x)[/tex] where [tex]y_1(x) \neq 0[/tex]


show that [tex]y_2(t) = y_1(t)\int_{t_0}^{t} \frac{1}{y_1(s)^2} e^{-\int_{t_0}^s p(r) dr} ds[/tex] is also a solution.

Homework Equations



I know that wronskian for this is defined

[tex]w(y_1,y_2)(t) = c \cdot e^{-\int p(t) dt}[/tex] according to Abels identity theorem.

The Attempt at a Solution



My question do I use this property of theorem and properties of the Wronskian to derive c?
 
Last edited:

Answers and Replies

  • #2
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Susanne -

The solution specified for y2 is the general form of a second solution to second order differential equation when the roots to its characteristic equation are repeated. I am not sure what to do when you have functions as coefficients, but perhaps if you can show your second order equation can be reduced to a second order equation with constant coefficients (with repeated roots), then you may assume the solution y2 = F(t)*y1 (F a function to be determined). Twice differentiating y2 and substituting it back into your original second order equation, you should obtain a new equation in terms of F', and F''. Hence you may let g = F' and obtain a first order, separable equation. Solve for g, then back-substitute F' and integrate to solve for F. Finally, since y2 = F(t)*y1, and since F(t) is known, you should get the answer specified for y2. Hope this helps.

- Sam
 
  • #3
317
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Susanne -

The solution specified for y2 is the general form of a second solution to second order differential equation when the roots to its characteristic equation are repeated. I am not sure what to do when you have functions as coefficients, but perhaps if you can show your second order equation can be reduced to a second order equation with constant coefficients (with repeated roots), then you may assume the solution y2 = F(t)*y1 (F a function to be determined). Twice differentiating y2 and substituting it back into your original second order equation, you should obtain a new equation in terms of F', and F''. Hence you may let g = F' and obtain a first order, separable equation. Solve for g, then back-substitute F' and integrate to solve for F. Finally, since y2 = F(t)*y1, and since F(t) is known, you should get the answer specified for y2. Hope this helps.

- Sam

does this have any relation to reduction of order tecnique ?
 
  • #4
162
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Yes, I believe the name of the method I described above is reduction of order. Sorry, I forgot its name.
 
  • #5
317
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Yes, I believe the name of the method I described above is reduction of order. Sorry, I forgot its name.

Why this isn't in my Calculus book. I don't know :(

Found this is a different textbook, but thanks ;)

Found this note here:

http://en.wikipedia.org/wiki/Reduction_of_order

You seem to me more intelligent than me and therefore is the ansatz mentioned

[tex]y_2(x) = y_1(x) \cdot v(x) [/tex] can this be describe as linearcombination of first solution and the abtr function v(x)??
 

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