can find C(x): We only need a single particular solution, so we arbitrarily select for simplicity. Therefore the particular solution is: The final solution of the differential equation is: This recreates the method of integrating factors. Specific second-order equation Let us solve We want to find the general solution to the differential equation, that is, we want to find solutions to the homogeneous differential equation The characteristic equation is: Since is a repeated root, we have to introduce a factor of x for one solution to ensure linear independence: and . The Wronskian of these two functions is Because the Wronskian is non-zero, the two functions are linearly independent, so this is in fact the general solution for the homogeneous differential equation (and not a mere subset of it). We seek functions A(x) and B(x) so A(x)u1 + B(x)u2 is a particular solution of the non-homogeneous equation. We need only calculate the integrals Recall that for this example That is, where and are constants of integration. General second-order equation We have a differential equation of the form and we define the linear operator where D represents the differential operator. We therefore have to solve the equation for , where and are known. We must solve first the corresponding homogeneous equation: by the technique of our choice. Once we've obtained two linearly independent solutions to this homogeneous differential equation (because this ODE is second-order) — call them u1 and u2 — we can proceed with variation of parameters. Now, we seek the general solution to the differential equation which we assume to be of the form Here, and are unknown and and are the solutions to the homogeneous equation. (Observe that if and are constants, then .) Since the above is only one equation and we have two unknown functions, it is reasonable to impose a second condition. We choose the following: Now, Differentiating again (omitting intermediary steps) Now we can write the action of L upon uG as Since u1 and u2 are solutions, then We have the system of equations Expanding, So the above system determines precisely the conditions We seek A(x) and B(x) from these conditions, so, given we can solve for (A′(x), B′(x))T, so where W denotes the Wronskian of u1 and u2. (We know that W is nonzero, from the assumption that u1 and u2 are linearly independent.) So, While homogeneous equations are relatively easy to