EXAMPLE 1 Illustration of Theorem 2 The functions and are solutions of . Their Wronskian is . Theorem 2 shows that these solutions are linearly independent if and only if . Of course, we can see this directly from the quotient . For we have , which implies linear dependence (why?). EXAMPLE 2 Illustration of Theorem 2 for a Double Root A general solution of on any interval is . (Verify!). The corresponding Wronskian is not 0, which shows linear independence of and on any interval. Namely, . A General Solution of (1) Includes All Solutions This will be our second main result, as announced at the beginning. Let us start with existence. THEOREM 3 Existence of a General Solution If p(x) and q(x) are continuous on an open interval I, then (1) has a general solution on I. PROOF By Theorem 1, the ODE (1) has a solution on I satisfying the initial conditions and a solution on I satisfying the initial conditions The Wronskian of these two solutions has at the value Hence, by Theorem 2, these solutions are linearly independent on I. They form a basis of solutions of (1) on I, and with arbitrary is a general solution of (1) on I, whose existence we wanted to prove.