# Higher-Order Linear Equations

• Oct 26th 2009, 07:02 PM
Higher-Order Linear Equations
Determine whether the following sets of functions are linearly independent for all x where the functions are defined. {(x^3)-3,(x^3)-3x,x-1}

I have no idea where to begin : (
• Oct 26th 2009, 07:05 PM
TheEmptySet
Quote:

Determine whether the following sets of functions are linearly independent for all x where the functions are defined. {(x^3)-3,(x^3)-3x,x-1}

I have no idea where to begin : http://mathhelpforum.com/differential-equations/110723-higher-order-linear-equations.html#post392216\" rel=\"nofollow\">
Determine whether the following sets of functions are linearly independent for all x where the functions are defined. {(x^3)-3,(x^3)-3x,x-1}

I have no idea where to begin : (

Do you know the definition of linear independence?

Functions, \$\displaystyle f_1, f_2, f_3, ... , f_n \$ are said to be linearly dependent if they can be written in the form:

\$\displaystyle A_1 f_1 + A_2 f_2 + A_3 f_3, ... , A_n f_n = 0\$, where \$\displaystyle A_1, A_2, A_3, ..., A_n \$ are non-zero constants.

So to prove that your three functions are linearly independent, you must show that there are NO non-zero constants, A, B and C for which the following is true:

\$\displaystyle A(x^3 - 3) + B(x^3 - 3x) + C(x-1) = 0\$

From here, I suggest expanding the brackets, then comparing the co efficients of \$\displaystyle x^3 \$, \$\displaystyle x^1 \$ and \$\displaystyle x^0 \$ on the RHS and LHS (So, for examples, the coefficients of \$\displaystyle x^3 \$ on the LHS is A+B, on the RHS is 0, therefore \$\displaystyle A+B = 0 \$). Hopefully you'll get three equations in A, B and C, and if your functions are truly linearly independent, then there will be no real non-zero solutions for those equations.
• Oct 26th 2009, 07:22 PM
Mush
Quote:

Im just not sure how you apply it though : / its so confusing

The Wronskian is pretty easy to use.

You start with n functions. You then construct an n x n matrix. In your first row, you put your functions. In the second row, you put the first derivative of your functions. In the third row, you put the third derivative of your functions. All the way down to the nth row which has the (n-1)th derivative. You then take the determinant of your matrix. If the determinant is 0, the functions are not linearly independent. If the determinant is non-zero at SOME point, x, then the equations are linearly independent.

So for you, you have 3 equations. So you're matrix will be a 3x3. So you'll need your functions for the first row, their first derivatives for your 2nd row, and their 2nd derivative for your 3rd row.

Once you're placed all these in the matrix, take its determinant, and if the result is 0, they are linearly dependent, if the result is non-zero for some x, then the are linearly independent.

I think, for this particular example, my method is a bit less work than using the Wronskian. But the Wronskian is better generally, and it's very easy to construct and solve as long as you don't go beyond 3x3 matrices!
• Oct 27th 2009, 10:48 PM