In general, for an nth order linear differential equation, if n-1 solutions are known, the last one can be determined by using the Wronskian. Consider the second order differential equation in
Lagrange's notation: y'' = a(x)y' + b(x)y where a(x), b(x) are known, and y is the unknown function to be found. Let us call y_1, y_2 the two solutions of the equation and form their Wronskian W(x) = y_1 y'_2 - y_2 y'_1 Then differentiating W(x) and using the fact that y_i obey the above differential equation shows that W'(x) = a(x) W(x) Therefore, the Wronskian obeys a simple first order differential equation and can be exactly solved: W(x) = C~e^{A(x)} where A'(x)=a(x) and C is a constant. Now suppose that we know one of the solutions, say y_2 . Then, by the definition of the Wrońskian, y_1 obeys a first order differential equation: y'_1 -\frac{y'_2}{y_2} y_1 = -W(x)/y_2 and can be solved exactly (at least in theory). The method is easily generalized to higher order equations. The relationship between the Wronskian and linear independence can also be strengthened in the context of a differential equation. If we have n linearly independent functions that are all solutions of the same monic nth-order homogeneous-linear ordinary differential equation y^{(n)}+Ly=0 (where L is a linear differential operator with respect to x of order less than n) on some interval I, then their Wronskian is zero
nowhere on I. Thus, counterexamples like x^2 and x (whose Wronskian is zero everywhere) or even x^2 and 1 (whose Wronskian 2x is zero somewhere) are ruled out; neither pair can consist of solutions to the same second-order differential equation of this type. (It's true that x^2 and 1 are both solutions to the same
third-order differential equation y^{(3)}=0. But the Wronskian -2 of the
three independent solutions x^2, x, and 1 is nowhere zero.) == Generalized Wronskians ==