I have introduced the notion of the fundamental set of the solutions, and proved to that, any given set of the homogeneous linear systems of equations always has a fundamental set of the solutions. Now it's time to discuss what's the meaning of the fundamental set of solution. The general solutions for homogeneous system. Consider the homogeneous system of equations say X prime is equal to A of X. Then let the X of J. For J equal to 1 to n be a fundamental set of solutions. Solutions for the problem on I for the given system of equations on I. Let's make the linear combination X. Summation of J equal to 1 through the n. CJ of X of T. Then, my conclusion is, this is a general solution of the system, X prime is equal to H of X on I. In other words, what I mean by a general solution is now that was for any solution X of the problem, X prime is equal to A times of X. For any given solution capital X, they always exist as the end constant. Suitable constants, CJ. Such that the end solution must be a linear combination with CJ of XJ's. In other words, if you consider this linear combination where the CJ is a arbitrary constant and consider this set of solutions. This is the set of all possible solutions. Solutions of the problem X prime is equal to A's of X. That's what I mean by saying that this is the general solution. Any solution must be of this form. Any solution of this problem must be of this form for suitable choice of the constant CJ. For example, in the last class, in the last lecture, I claim that X1 equal to 1 minus 1 E to the T. X2 is equal to 1 3 E to the T, and plus 4 minus 4 E to the 2 T. These two is a fundamental set of solutions. Solutions for the problem x prime is equal to, let me remind it. Let me see. Is the fundamental set of solutions for the problem minus 1 and 0 and x, y. In the example we handled in the last lecture, we showed that these two are linearly independent solution of this system so that this former fundamental set of the solutions. What can we say then? General solution of this problem. General solution of this system is x of t is equal t just a linear combination of these two, X1 and X2. I see 1 minus 1 e to the t plus c_2 and 1,3 to the t plus 4 Negative 4 times t, e to the t. This is a general solution. In other words, the collection of all those expressions with arbitrary constant C_1 and C_2, they form a full set of all possible solutions of the given system. Now up to this point we consider only the homogeneous linear system of equations. As we did for any single linear differential equations in the Part 1. Now it's time to consider the non-homogeneous system of equations. We'd like to discuss things for the non-homogeneous problem. This is the homogeneous system of equations. If we add some non-homogeneous vector F capital F, then this is possibly non-homogeneous linear system of equations. Non-homogeneous system. We're interested in a general solution of this non-homogeneous system. Now here's the theorem. General solution of the non-homogeneous system. Consider the n is such a non-homogeneous system of equations and a reminder that as before, we'll always assume that the entries of a and the entries of capital F, they are continuous functions on a common interval I, so think about such a non-homogeneous system of equations. I claim that our general solution, in other words, the set of all possible solutions. A general solution of this non-homogeneous problem is x is equal to x sub c plus x sub p. What is x sub c? This is the so-called the complimentary solution, which remains a general solution of corresponding homogeneous problem, x prime is equal to a times x. That is the complimentary solution. What is x of p? X sub p, this is any particular solution of the given non-homogeneous problem, say x prime is equal to a times x plus capital F. The proof of this theorem is easy enough. If you remind the proof of the same powerful theorem for any single linear or the non-homogeneous equation, we have a such a theorem. The corresponding theorem for a single equation and the proof of this fact is exactly the same as the proof of that one. Proof of the same type of theorem for linear single non-homogeneous equation so I will not repeat the proof here again. Again, why we call such a form to be a general solution of this non-homogeneous problem. It means any solution of this problem must be of this form for suitable choice of the constant, c_js appearing in this complimentary solution. To be precise, I mean, is this one, what is X_c? X_c must be a linear combination of any fundamental set of solutions for corresponding homogeneous problem. X_c is of the form c_j, X_j. Then plus X_p. What is X_j's? This is a fundamental set of solutions for the corresponding homogeneous problem. In other words, these are n linearly independent solutions of X prime is equal to A times of X. Let me repeat my claim again. We call this form of the solution to be a general solution of this problem. Because any solution of this problem must be of the form this for suitable choice of c_1 and c_2 and so on, c's of n. That's really the meaning behind the general solution. One simple example. I'm considering the non-homogeneous problem, X prime is equal to 2, 1 minus 1, 0 and the X, and we have a non-homogeneous term, minus 3t minus 3 and the 2t. Recognize that if there is no such term, know the non-homogeneous term. Look at only this part of first. That's the problem that we have considered before, homogeneous problem. It has a two linearly independent solution X_1 is equal to 1 minus 1e^ t, and X_2 is equal to 1, 3 e^t and plus 4 minus 4 and the te^t. These are the two linearly independent solution of the corresponding homogeneous problem. Now we are interested in solving this, the non-homogeneous problem. Then we say that general solution is equal to X_c plus X_p. We know that. What is X_c? That's arbitrary linear combination of these two. C_1, X_1 plus c_2X_2 and the plus any particular solution. We do not know this particular solution yet. It's just, I claim that then. Think about the vector 2t plus 1, and the minus t plus 3. Plugging this expression into this differential equation. You can confirm that, it satisfy the given differential equation. It's easy computation. Do not ask how I find that this is a solution. We will discuss such a problem a little bit later. I claimed that this is a solution. This is a particular solution of the given non-homogeneous problem. That means what? By the theorem, we have introduced c_1 X_1 plus c_2 X2 plus X_p. That is a 2t plus 1 and minus t plus 3. This is the general solution of our original problem. Where the X_1 and the X_2 are the same as here.