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

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
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 $(0,1)$.