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Differential equations - linear dependency

  • #1
162
0
Let y1(t) = t^2 and y2(t) = t|t|
(a) Show that y1 and y2 are linearly dependent on the interval 0<=t<=1.
(b) Show that y1 and y2 are linearly independent on the interval -1<=t<=1.
(c) Show that the Wronskian W[y1,y2] is identically zero.



My attempt:
(a) So I know that if the Wronskian is zero, then y1 and y2 are linearly dependent. So i worked out the wronskian and i got
w[y1,y2](t) = t^3 - t^2|t|
I'm not sure how to show that it is linearly dependent on the interval. should i substitute in the intervals 0 and 1 and find that they are equal to zero? what about everything inbetween 0 and 1?

(b) So for linearly independence, wronskian doesn't equal zero, again i have the same question of whether i substitute in the interval, and what do i do with the numbers inbetween, i am unsure how to prove this.

(c) i have no idea what identically zero means, can someone explain this?
 

Answers and Replies

  • #2
78
0
Is that all the information they gave you? Did they give you a differential equation for the solutions?
 
  • #3
162
0
yes, this is all. and i think i have solved (c). i did the wronskian wrong in (a), when i did it again , i got exactly 0, and i guess this is what (c) is asking.

but for (a) and (b), can i just show that for dependence, y1 = k y2, k should exist, and for independence, k should not exist? is that sufficient.
 
  • #4
HallsofIvy
Science Advisor
Homework Helper
41,808
933
If t is non-negative, as it is on [0, 1], then t|t|= t2. The two functions you give are not "independent", they are exactly the same!

"Identically zero" means equal to 0 for all values of t.

but for (a) and (b), can i just show that for dependence, y1 = k y2, k should exist, and for independence, k should not exist? is that sufficient.
For two functions, yes, but for more than two you will need to use the general definition of independence: If
[tex]k_1 f_1(x)+ k_2 f_2(x)+ \cdot\cdot\cdot + k_n f_n(x)= 0[/tex]
for all x with an least one ki not equal to 0, then the functions are dependent.
 

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