The Wronskian and the Derivative of the Wronskian

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SUMMARY

The discussion focuses on the Wronskian and its derivative in the context of the second-order linear differential equation \(y'' + ay' + by = 0\), where \(a\) and \(b\) are non-zero real numbers. Participants clarify that the Wronskian \(w(y_1, y_2)\) is calculated as the determinant of the matrix formed by the functions and their derivatives, specifically \(w(y_1, y_2) = y_1y_2' - y_2y_1'\). It is established that if the initial vectors \((y_1(0), y_1'(0))^T\) and \((y_2(0), y_2'(0))^T\) are linearly independent, then \(y_1(t)\) and \(y_2(t)\) are also linearly independent functions.

PREREQUISITES
  • Understanding of second-order linear differential equations
  • Familiarity with the concept of the Wronskian
  • Knowledge of linear independence of vectors
  • Basic calculus, specifically differentiation
NEXT STEPS
  • Study the properties of the Wronskian in differential equations
  • Learn how to apply the Wronskian to determine linear independence
  • Explore examples of second-order linear differential equations
  • Investigate the implications of the Wronskian in the context of solutions to differential equations
USEFUL FOR

Students and educators in mathematics, particularly those studying differential equations, as well as researchers interested in the properties of linear independence in function spaces.

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Homework Statement



y1(t) and y2(t), 2 solutions of the equation:

y'' +ay'+by=0, with a,b εℝ - {0}
a) Determine:
d/dt w(y1,y2)
where w(y1,y2) is the wronskian of y1(t) and y2(t)
b)
Deduce that if (y1(0),y1'(0)^T and (y2(0), y2'(0))^T are 2 linearly independent vectors. Then y1(t) and y2(t) are linearly independent functions.

Homework Equations


^T = transpose
the wronskian is the det |y1 y2|
|y1' y2'| = y1y2' -y2y1'
Vectors are linearly independent if w(y1,y2) does not equal 0

The Attempt at a Solution


For part a, do I just find the wronskian of y1 and y2 and then take the derivative?
For part b I'm super confused. I notice that if you transpose the two vectors and put them into a determinant than they are the wronskian.. other than that I'm pretty lost..
 
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for a. yes, although i think you can simplify the result a little...

for b. start with the given linear combination equal to zero...
 
I'm a bit confused regarding the linear combination. Would it just be

C1V1+C2V2=0?
I'm not sure what this would accomplish... other than c1v1=-C2V2...
 

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