Compute the Wronskian & Simplify

  • Context: MHB 
  • Thread starter Thread starter shamieh
  • Start date Start date
  • Tags Tags
    Simplify Wronskian
Click For Summary
SUMMARY

The discussion focuses on computing the Wronskian for the functions $y_1 = t^2 + 1$ and $y_2 = 3t^2 + k$, resulting in $W(y_1, y_2) = 6t - 2kt$. The functions are confirmed to be linearly independent when the Wronskian is not identically zero, which occurs for all values of $k$ except $k = 3$. This conclusion is reached by analyzing the Wronskian and the differentiability of the functions.

PREREQUISITES
  • Understanding of Wronskian computation
  • Knowledge of linear independence in the context of functions
  • Familiarity with differentiation of polynomial functions
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of the Wronskian in detail
  • Explore the concept of linear independence in vector spaces
  • Learn about the implications of differentiability on function behavior
  • Investigate examples of Wronskian applications in differential equations
USEFUL FOR

Students and educators in mathematics, particularly those studying differential equations and linear algebra, as well as anyone interested in the application of the Wronskian in determining function independence.

shamieh
Messages
538
Reaction score
0
Compute the Wronskian and simplify.
So the first part was easy

a) $y_1 = t^2 + 1$ , $y_2 = 3t^2 + k$

=$6t-2kt$

b) for what values of $k$ are the functions linearly independent

so would I just solve for $k$? I'm confused

What exactly does linearly independent mean?
 
Physics news on Phys.org
Ok so I figured out it is Linearly independent because after I took the Wronskian I didn't get 0. But for what values of $k$ ? Would it just be as long as $k > 0$ ?
 
You can use the Wronskian to show linear independence as follows: if the Wronskian is not identically zero, and the two functions are (infinitely) differentiable, then the two functions are linearly independent. For what values of $k$ does this happen?
 
Ackbach said:
You can use the Wronskian to show linear independence as follows: if the Wronskian is not identically zero, and the two functions are (infinitely) differentiable, then the two functions are linearly independent. For what values of $k$ does this happen?
as long as $k$ > 0 ?
 
the two functions are inifnitely differentiable

because $y_1 = t^2 + 1 $

'= 2t + 1
'' = 2
''' = 0
''''=0
...etc

and $y_2 = 3t^2 + k$
' = 6t
'' = 6
''' = 0
etc

So would it be all values as long as $k \ne 3$
 
Last edited:
wait so how about

t(6-2k) when k = 3 then t(6-6) = 0?

so would it be k can be any value except when k is 3?
 
shamieh said:
wait so how about

t(6-2k) when k = 3 then t(6-6) = 0?

so would it be k can be any value except when k is 3?

You got it!
 

Similar threads

Replies
2
Views
1K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
6K
  • · Replies 1 ·
Replies
1
Views
2K