# Recitation 8

Example 1 Determine whether 2x - 3, x^3 + 1, 2x^2 - 1, x^2 + x + 1 are linearly independent.

Solution The Wronskian of 2x - 3, x^3 + 1, 2x^2 - 1, x^2 + x + 1 is \begin{vmatrix}2x - 3 & x^3 + 1 & 2x^2 - 1 & x^2 + x + 1 \\ 2 & 3x^2 & 4x & 2x+1 \\ 0 & 6x & 4 & 2 \\ 0 & 6 & 0 & 0\end{vmatrix} = -24 \neq 0. Therefore these functions are linear independent.

Remark Wronskian is defined for any set of functions. There functions do not necessarily come from ODEs.

Strategy Find n solutions y_1, \dots, y_n to an nth order homogeneous linear differential equation and compute W(y_1, \dots, y_n). If W(y_1, \dots, y_n) \neq 0, then y_1, \dots, y_n form a fundamental set of solutions, that is, the general solution is y = c_1y_1 + \dots c_n y_n.

Example 2 Verify 1, x, x^3 form a fundamental set of solutions of xy''' - y'' = 0.

Solution It is easy to verify 1, x, x^3 are three solutions of xy''' - y'' = 0. Their Wronskian is given by \begin{vmatrix}1 & x & x^3 \\ 0 & 1 & 3x^2 \\ 0 & 0 & 6x\end{vmatrix} = 6x \neq 0. This shows they form a fundamental set of solutions.

Example 3 Use the method of reduction of order to solve (2-x)y''' + (2x-3)y'' - xy' + y = 0 given that y_1 = e^x is a solution.

Example 4 Use reduction of order to solve x^2(x+3)y''' - 3x(x+2)y'' + 6(1+x)y' - 6y = 0 given that y_1 = x^2 is a solution.