Sachchidanand Prasad

10-03-2023

Problem: Show that the solution $e^x, e^{-x}, e^{2x} $ of $y^{\prime \prime\prime } - 2 y^{\prime\prime} -y^\prime +2y = 0$ are linearly independent.

Solution: In order to show that the functions are linearly independent, we will consider the Wronskian. We recall that the Wronskian for functions $f_1(x),\cdots, f_n(x)$, $W \left( f_1,f_2,\ldots ,f_n \right)(x)$, is the determinant of the following matrix. \[ \begin{pmatrix} f_1(x) & f_2(x) & \cdots & f_n(x) \\ f_1^\prime (x) & f_2^\prime (x) & \cdots & f_n^\prime (x) \\ \vdots & \vdots & \vdots & \vdots \\ f_1^{(n-1)}(x) & f_2^{(n-1)}(x) & \cdots & f_n^{(n-1)}(x) \end{pmatrix}. \]

Let $f_1,f_2,\ldots,f_n$ be differentiable on $[a,b]$. If Wronskian is nonzero for some $x_0 \in [a,b]$ then functions are linearly independent on $[a,b]$. If $f_1,\ldots ,f_n$ are linearly dependent, then the Wronskian is zero for all $x\in [a,b]$.

Note that the Wronskian for the given functions will be \begin{align*} W\left( e^x, e^{-x}, e^{2x} \right) & = \begin{vmatrix} e^x & e^{-x} & e^{2x} \\ e^x & -e^{-x} & 2e^{2x} \\ e^x & e^{-x} & 4e^{2x} \\ \end{vmatrix} \\ & = e^x \begin{vmatrix} -e^{-x} & 2e^{2x} \\ e^{-x} & 4e^{2x} \end{vmatrix} - e^{-x} \begin{vmatrix} e^x & 2e^{2x} \\ e^{x} & 4e^{2x} \end{vmatrix} + e^{2x} \begin{vmatrix} e^x & -e^{-x} \\ e^{x} & e^{-x} \end{vmatrix} \\ & = e^x \left( -4e^x - 2e^x \right) - e^{-x} \left( 4e^{3x} - 2e^{3x} \right) + e^{2x} (1+1) \\ & = -6e^{2x} - 2e^{2x} + 2e^{2x} \\ & = -6e^{2x}. \end{align*} Here the Wronskian is non-zero for $x=0$ (it is nowhere zero, but this is not required) and hence the given solutions are linearly independent.

We can write a general solution to the given differential equation. As the equation is of third degree and there are three linearly independent solutions, so the general solution of the differential equation will be \[ c_1e^{x} + c_2e^{-x} + c_3 e^{2x}, \] where $c_1,c_2$ and $c_3$ are constants.