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?

Show Answer
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

## Wronskian

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 \(y^{''} + 5y^{'} - 50 y = 0 \)

Show Answer
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.