RF's question at Yahoo Answers (linear independence, Wronskian).

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SUMMARY

The discussion addresses the determination of linear independence between the functions y1(t) = te^(2t) and y2(t) = e^(2t) using the Wronskian method. The Wronskian is calculated as W(y1, y2)(t) = -e^(4t), which does not vanish identically on the interval (0, 1). Therefore, it is concluded that the functions y1 and y2 are linearly independent on this interval.

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  • Understanding of differential equations
  • Familiarity with the concept of linear independence
  • Knowledge of the Wronskian determinant
  • Basic calculus, including differentiation
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Students and educators in mathematics, particularly those focusing on differential equations, linear algebra, and calculus. This discussion is beneficial for anyone seeking to understand the relationship between functions in terms of linear independence.

Fernando Revilla
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Here is the question:

So I have a question from the topic of differential equations about linear independence/Wronskian...

The problem states: In this problem, determine whether the functions y1 and y2 are linearly dependent on the interval (0, 1).

y1(t) = te^(2t), y2(t) = e^(2t)

Please explain how to do this problem step by step because I have no clue what to do...
Thank you!

Here is a link to the question:

Differential Equations...Linear independence question? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
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Hello RF,

For $n$ functions $y_1(t), y_2(t) , \ldots , y_n(t)$ which are $n-1$ times differentiable on an interval $I$, the Wronskian is defined by:
$$W(y_1, \ldots, y_n) (t)=
\begin{vmatrix}
y_1(t) & y_2(t) & \cdots & y_n(t) \\
y_1'(t) & y_2'(t) & \cdots & y_n' (t)\\
\vdots & \vdots & \ddots & \vdots \\
y_1^{(n-1)}(t)& y_2^{(n-1)}(t) & \cdots & y_n^{(n-1)}(t)
\end{vmatrix}\quad (t\in I)$$
In our case,
$$W(y_1, y_2) (t)=
\begin{vmatrix}
y_1(t) & y_2(t) \\
y_1'(t) & y_2'(t)
\end{vmatrix}=\begin{vmatrix}
te^{2t} & e^{2t} \\
(1+2t)e^{2t} & 2e^{2t}
\end{vmatrix}=-e^{4t}\quad (t\in (0,1))$$
According to a well-known property, the functions are linearly independent on $I$ if the Wronskian does not vanish identically. Clearly, this condition is satisfied, so $y_1(t),y_2(t)$ are linearly independent on $(0,1)$.
 

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