Now, let's consider the one more example. We'd like to consider the problem, X prime is equal to minus 2, 1 and 1 minus 2 and the X and plus 2_e to the minus t minus e to the minus t minus 4 and 5. You're considering this problem. Find its general solution. You can write it as AX plus, decompose it. This is 2, 0 e to the minus t plus minus 4, 5. The non-homogeneous term involves constant vector and the same constant vector times e to the minus t. A has an eigenvalue A. This is A. A has an eigenvalues minus the 1 and the minus 3 with corresponding eigenvectors, say, 1, 1 and 1, minus 1. That we can write immediately the complimentary solution. This is a C 1, 1 1, e to the minus t, and plus C 2, 1 minus 1, e to the minus 3_t. That's the complimentary solution. Again, is time to make a reasonable guess for the particular solution. Looking at these non-homogeneous terms, it contains a constant term and the e to the minus t term, so our reasonable guess is, for the particular solution X, X_p is equal to undetermined coefficients A_1, A_2, e to the minus t, and the plus another undetermined coefficients, I'll write it as a C_1 and C_2. This is our first guess. Then you have to compare this our guests, initial guess X sub p and this complimentary solution. E to the minus t times the constant vector is a part of this complimentary solution. Our initial guess also contains e to the minus t term times the some undetermined coefficients. You may think that, this guess is no good, but we should make the modification, say, A_1 A_2, t times e to the minus t. That is something we have learned from our experience for the singer linear non-homogeneous equation when we would like to make a guess for the particular solution X_p. Modify it by this way, t times the e plus from this constant term to 0. We have to still have a, C_1, C_2. If we do the computation, actual computation with this guess, you can see that immediately there, look through this little bit computation, it does not work. It fails to give a particular solution. My claim is, you need term in between on other undetermined coefficients B_1 B_2 times e to the minus t. That's the difference. We need this term. That's the difference between the case of the single non-homogeneous equation and the system of equations. It'll be a good exercise to see that actually why this initial guess or the second guess without the terms in the middle phase. Now make this top of the modification, plugging this X_p into the differential equation. You are going to get six simultaneous a system of equations as a user. I'm skipping the actual computation plugging this one into that. For the six unknowns, a_1, a_2, and b_1, b_2 and c_1, c_2. If we do that, then you will get. You should have carry out the actual computation. This is equal to 1, 1. b_1, b_2 is equal to b_1 and b_1 minus 1, where the b_1 is arbitrary. Finally, c_1, c_2 is equal to minus 1 and 2. By plugging this X sub p into the differential equation, set the equality between the left hand side and the right-hand side, so those they obtain the simultaneous equations. You can conclude this. What is the mean then? X sup p s equal to a_1, a_2, 1, 1, t times e to the minus t. This is the guessed. Plus b_1, b_2, that is equal to b_1, b_1 minus 1 e to the minus t plus c_1, c_2 is equal to minus 1 and 2. This is the particular solution we obtained where the b_1 is arbitrary. Next we'll look at this one again. You can write it as a 1, 1 t times e to the minus t. From this, you will get, you can decompose it as a b_1, 1, 1 e to the minus t plus 0 minus 1, e to the minus t plus minus 1 and the 2. You can rely to decompose it into that way. Note the following. Notice that, pay attention to this term. Let's think about this term. Arbitrary equals the b_1 times a constant vector 1, 1, e to the minus t. Look at this complimentary solution. There we have arbitrary equals c_1 times vector 1, 1 times e to the minus t. This is exactly. This part is exactly. This is a part of complimentary solution where the b_1 is arbitrary. Then what is that mean then? You can choose b_1 to be any real number, maybe, in some sense, the best choice is take b_1 equal to 0 because forming this general solution by X_c plus X sub p. This arbitrary equals times 1, 1 e to the minus t is already a part of the X_c. It will be absorbed into that term so that by taking b_1 equal to 0, still we have X sub c and plus 1, 1 t e to the minus t plus 0 minus 1 e to the minus t plus minus 1 and 2. This is a typical general solution of this original problem.