So, fortunately we have two equations where the two unknowns are u_1 and u_2, right? What are they? First one is this one. Require u_1'y_1'+u_2'y_2'=0. This is one such equation. Now, we have another equation for the unknowns u_1 and u_2, right here. u_1'y_1'+u_2'y_2'=0. We have two unknowns, u_1 and u_2, and we have two equations, u_1'y_1+u_2'y_2=0 and u_1'y_1'+u_2'y_2'=g. That's what we are expecting in fact. What I mean is, therefore, we obtain a system of equations with two unknowns, u_1 prime and u_2 prime. Say, y_1u_1'+y_2u_2'=0. Another one is, y_1'u_1'+y_2'u_2'=g. We have two unknowns, u_1_prime and u_2_prime and we have two equations. This is a system of equations for two unknowns and linear, a linear system of equations for two unknowns, u_1_prime and u_2_prime. Apply the Cramer's rule. By the Cramer's rule for solving this linear system of equation then, first unknown u_1 prime is equal to determinant of the coefficient matrix. What is the determinant of the coefficient matrix? Coefficient matrix is y_1, y_2, and y_1 prime, and y_2 prime. We need this determinant, right? Can you remind what it is? We use this one as a W. This is the so called the Wronskian of two linearly independent functions y_1 and y_2, where W of y_1 and y_2, which I simply write by capital W. Its determinant is equal to y_1, y_2 prime minus y_2, y_1 prime. So, by the Cramer's rule, the first unknown u_1_prime is equal to determinant of two by two matrix over determinant of the coefficient matrix which is the Wronskian. So, in fact, this will be u_1_prime is equal to minus y_2 times G over W. On the other hand, the second unknown u_2_prime is equal to determinant of y_1 zero, y_1_prime g over the Wronskian, which is equal to y_1 times g over W. So, in a sense, we solve the two unknowns, u_1_prime and u_2_prime. u_1 and u_2, we can obtain by solving this simple first-order differential equation or simply through the integration. What is u_1? u_1 must be antiderivative of this part. And what is u_2? u_2 must be the antiderivative of this expression. Obtaining u_1 and u_2 in this way, due to this u_1 and u_2, and make the final Y_P. This is u_1, y_1 plus u_2, y_2,. y_1 and y_2, they are given or they are known already; u_1 and u_2, you can obtain by solving by integrating the equation four. Using those four functions for y of p, you get a particular solution of the original problem. So, the general solution of the original problem, general solution to y double prime plus p y_prime plus qy is equal to g. Now, you can get y is equal to, y_c is what? c_1 y_1 plus c_2 y_2 and plus y_p. This is y of c, complementary solution. y of p, that is equal to u_1 y_1 plus u_2 y_2. This is a particular solution y of p. So, we are done. We call this process of obtaining a particular solution of this form for the nonhomogeneous differential equation, we call it the variation of parameters because you are replacing arbitrary constant c_1 and c_2 by unknown function u_1 and u_2. In summary, we can state, we can have the following theorem, the variation of parameters. Assume that all the functions involved, say, p of x and q of x and g of x, they are continuous in certain interval from alpha to beta, and assume that y_1 and y_2 are two linearly independent solutions of corresponding homogeneous equation, say, y double prime plus p(x) y_prime plus q(x) y is equal to zero, then, the particular solution is given by y_1 times u_1. What is u_1? Antiderivative of negative y_2 g over W plus y_2 times u_2. What is u_2? Antiderivative of y_1 g over W. Can you remind it? I'll show it over here. u_1 prime is equal to negative y_2 g over W. On the other hand, u_2 prime is equal to y_1 g over W. So, u_1 is an antiderivative of negative y_2 g over W and u_2 is antiderivative of y_1 g over W. That's what I'm writing here. So, y_1 times u_1 and y_2 times this is u_2 part, where the capital W, this is Wronskian of two linearly independent function y_1 and y_2. So this is the required particular solution of our original homogeneous problem. So finally, as I said here, the general solution of our original problem is y_c plus y_p, and y_c is simply linear combination of y_1 and the y_2. So, we completely solve this possibly variable coefficient linear nonhomogeneous differential equation. I'll illustrate this theorem, the variation of parameters, through a couple of examples.
