Let's think about the one more example. We'd like to solve the following, the initial value problem of two-by-two systems. The problem is x prime is equal to the negative 4 and 2 minus 1 and times x, and plus the non-homogeneous term, that is a 2 over t minus 1 and 4 over t. Initial condition is given at the 0.1. X sub 1 is equal to negative 4 over 25 and 2 over 25. Before tackling the problem, let's note, the following. The entries of this non homogeneous term, they say the 2 over t minus 1 and the 4 over t. These two entries are continuous on the whole real line except the t is equal to 0 because if we have t in the denominator. In other words, the entries of this capital F of t, the non-homogeneous term, is the continuous on the interval from 0 to infinity, or the interval from minus infinity to 0. We're given the initial condition. Initial condition is given at time t is equal to 1. Then means we're supposed to solve the problem on the interval 0, infinity which contains the initial point to 1. Then let's go back to our original problem. Yeah, handling this one, we needed to find the force to find the complementary solution, in other words, the general solution of the corresponding homogeneous problem. Then we have to find the particular solution satisfying this non-homogeneous problem. Look at this, the entries of the non-homogeneous term again. Then it suggests to us that the method of undetermined coefficients is not applicable in this case. We have to use the method of variational parameters away. Anyway, let's try to solve this corresponding homogeneous problem. Look at this coefficient matrix a. Then the matrix a has two eigenvalues, eigenvalues 0 and minus 5, and each one of them has a corresponding eigenvectors. First, the 1 and 2, and the second one is a minus 2 and 1. That means what? We have two linearly independent solutions given by e to the 0t, that is one times this corresponding eigenvector. Second solution is e to the minus 5t times this corresponding eigenvector. These two are the two linearly independent solution of the corresponding homogeneous problem. That the so-called the fundamental matrix, capital V of t is the 1 and 2 and minus 2 times u to the minus 5t and the 1 times e to the minus 5t. This is the fundamental matrix. Yeah, then it's a straightforward to write down the particular solution, x_p. What is x_p then? The theory says, it is this fundamental matrix times anti-derivative of inverse of capital P of t and times the non-homogeneous vector. Say 2/t minus 1 and the 4/t, and the t is the anti-derivative. We need to compute the inverse of this fundamental matrix. Let's do it down here. Then, Inverse of the capital P, that is the 1 over determinant of capital P, a determinant is what? One times e to the negative 5t minus 2 times e to the minus 5t. That either via let me see, e to the minus 5t and minus this times that, that is minus 4 times e to the minus 5t. In fact we get plus 4 times e to the minus 5t, That's the determinant. Then we need to switch these two. E to the minus 5t and one down there, and take the opposite sign for these two parties. That is 2 times e to the minus 5t and minus 2, test the inverse. In fact that this determinant will be this plus that will be equal to 5 times e to the minus 5t. Or is better to write it as 1/5 times e^5t. That's the inverse of the capital P. Plugging this into that, then we get. The particular solution will be instead of this I will write, the inverse sub capital P that we found the 1/5 and e^5t over 5 times e to the minus 5t, 2 times e to the minus 5t and minus 2 times 1, and we need 2/t minus 1 and the 4/t and the t. That's the particular solution. It's a simple method to find this product right. Doing that, you will get 1/5. This two-by-two matrix times the column vector, that is the 10/t minus 1 and the 2 times e^5t and dt. What's the anti-derivative of each component? That is 1/5 capital P of t. Then we have what? The 10 log of t minus t plus arbitrary constant, which we can take to be zero. From this one we have 1/10 e^5t plus arbitrary integral constant. That again, we can take it to be zero. We get this one. Here we note the following; the anti-derivative of 1/t is in fact the log of absolute value of t, but as I said in the beginning, because of this initial condition, initial condition is given at t is equal to positive 1, so we are supposed to solve the problem on the positive real axis from 0 to infinity so that we don't need the absolute value sign for the log part, but I just simply say that 10 times of log of t. To this computation with the capital P then, I claim that this will be equal to 1/5 times 10 log of t minus t minus 4/5 and 20 log of t minus 2t, and plus 2/5. I made a simple mistake down here, the anti-derivative of 2 times e to the 5t is not 1/10 e to the 5t. This is a stupid mistake, let me correct it. This is 2/5 times e to the 5t and then computing this multiplication, then finally we get this one. Now we're ready to write down the general solution of the x of t of this problem then. What is the general solution? The general solution x of t is; this is one solution for the corresponding homogeneous problem. That is another linear independent solution of the corresponding homogeneous problem. Complimentary solution will be the linear combination, so plus c_2 e to the minus 5t and minus 2 and 1, and plus. Up to this point, this is so-called the complimentary solution. Let me erase this part. This is equal to so-called x_c. Now we need to add the particular solution is right there. This is 1/5 and 10 times the log of t minus t minus 4/5, and 20 log of t minus 2t plus 2/5. That's the general solution. This part is the particular solution, x_p. Now we're applying the initial condition, then it's a simple computation. Using the initial condition plug at t is equal to 1 on the both sides and the required to be minus 4/25 and 2/25, then, it gives c_1 is equal to 1/5 and c_2 is equal to 0. You can confirm it easily. Plugging those two numbers into this one, we have the solution for the initial value problem and this given by the 1/5, and 10 log of t minus t plus 1/5, and 20 log of t minus 2t and plus 12/5. That's a solution to this given initial value problem.