We have learnt that Superposition Principle allows us to find new solutions from linear combinations of known solutions. However, we may be interested in fundamental solution set which is a set of solutions that can be used as the building blocks to find all other solutions.
A fundamental set of solutions is a set of linearly indepenent solutions to a DE on an interval, \(I\). A linearly independent set of functions on interval \(I\) only has a trivial solution to the following equation:
\(c_1f_1(x) + c_2f_2(x) + \dots + c_nf_n(x) = 0\)
Are the functions \( x^2 \) and \(2x^2 \) linearly independent?
The linear combination of the functions is:
\(c_1x^2 + c_2 2x^2 = 0\)
This simplifies to:
\(c_1 + 2c_2 = 0\)
This equation has infinite solutions. We can take \(c_2 = 1\) which gives \(c_1 = -2\).
\(-2x^2 + (1)2x^2 = 0\)
\(-2x^2 + 2x^2 = 0\)
\(0 = 0\)
Thus, the equations \(x^2\) and \(2x^2\) are linearly dependent.
One method to determine if functions are linearly independent is to check the Wronskian. The Wronskian of a group of functions is defined as:
\(W(f_1, f_2, \dots f_n) = \begin{vmatrix} y_1 & y_2 & y_3 \\ y^{'}_1 & y^{'}_2 & y^{'}_3 \\ y^{''}_1 & y^{''}_2 & y^{''}_3 \end{vmatrix} \)
A set of functions is linearly independent on interval \(I\) iif:
\(W(f_1, f_2, \dots f_n) \ne 0 \)
Confirm the set of solutions \( \{e^{-10x}, e^{5x}\} \) is a fundamental solution set to the DE \(y^{''} + 5y^{'} - 50 y = 0 \)
We have shown in the previous lesson that both solutions are indeed solutions to the DE. Next, to confirm they form a fundamental solution set, we can check the Wronskian:
\( W(e^{-10x}, e^{5x}) = \begin{vmatrix} e^{-10x} & e^{5x} & \\ -10e^{-10x} & 5e^{5x} \\ \end{vmatrix}\)
\(=e^{-10x} \cdot 5e^{5x} - -10e^{-10x} \cdot e^{5x} = 5e^{-5x} + 10e^{-5x} = 15e^{-5x}\)
\(W(e^{-10x}, e^{5x}) = 15e^{-5x}\)
Since \(15e^{-5x} > 0\), the set is linearly independent and the set is a fundemental solution set.