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

In summary, the Wronskian is used to determine the linear independence of functions on a given interval. In this problem, the Wronskian of the given functions is non-zero, therefore the functions are linearly independent on the interval (0,1).
  • #1
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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  • #2
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 $(0,1)$.
 

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

What is linear independence in mathematics?

In mathematics, linear independence refers to a set of vectors where none of the vectors can be written as a linear combination of the others. This means that each vector in the set carries unique information and is not redundant. Linear independence is an important concept in linear algebra and is often used to solve systems of linear equations.

How is linear independence related to Wronskian?

The Wronskian is a mathematical tool used to determine the linear independence of a set of functions. It is a determinant that can be calculated using the derivatives of the functions in the set. If the Wronskian is non-zero, then the functions are linearly independent. If the Wronskian is zero, then the functions are linearly dependent.

What is the significance of linear independence in solving systems of equations?

Linear independence is crucial in solving systems of equations because it allows us to determine the number of solutions. For example, if the equations are linearly independent, then there will be a unique solution. If the equations are linearly dependent, then there will be infinite solutions or no solutions at all.

Can linearly dependent functions still be useful in mathematical applications?

Yes, linearly dependent functions can still be useful in mathematical applications. For example, in physics, dependent variables can be expressed as linear combinations of independent variables, which simplifies the equations and makes them easier to solve. However, it is important to note that dependent functions cannot be used to uniquely determine the values of the variables.

How can I determine if a set of vectors is linearly independent?

To determine if a set of vectors is linearly independent, you can perform the Wronskian test. Calculate the Wronskian using the derivatives of the vectors. If the determinant is non-zero, then the vectors are linearly independent. Another approach is to set up a system of equations using the vectors and solve for the coefficients. If there is a unique solution, then the vectors are linearly independent.

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