 We can also have non-homogeneous systems of differential equations. How do we solve those? Good question. It's helpful to introduce something called the fundamental matrix. Suppose a differential equation has a solution that can be written as a linear combination of vectors multiplied by some exponential function. The solution can be represented in terms of the fundamental matrix C, where C is equal to a matrix whose columns are the individual vectors. And so our solution, X, is going to be C times a column vector of the constants. So for example, let's find the fundamental matrix for the system derivative is equal to 3122 times X. So first we find the eigenvalues and the eigenvectors of the coefficient matrix, which means we could write down the general solution. Now while we ordinarily write the general solution in this form, if we write the general solution in matrix form we get, and if we think about this as the product of a matrix of constants times the fundamental matrix, we can recover the fundamental matrix. So remember in matrix multiplication, first row times first and only column gives us the first entry of the product. Since we want to get minus C1e to the t plus C2e to the 4t, that means our first row must be minus e to the t e to the 4t. Similarly, our second row times our column has to give us 2C1e to the t plus C2e to the 4t. And that means the second row must be, which gives us our fundamental matrix. So now we can talk about non-homogeneous equations. Suppose dx equals ax has solution cc, where c is a column vector of constants. Then dx equals ax plus g of t might have a solution cc of t, where c of t is a column vector of functions of t. Now all we have to do is figure out what c of t is. Now let's jump right in and try to solve this non-homogeneous system. Now we've already determined that the homogeneous system has solution. So we'll use variation of parameters and we'll suppose that we have a solution of the form almost the same thing, but instead of constants c1 and c2, we'll have functions of t, c1 and c2. So our differential equation requires us to find the derivative of x. Now since x is a product of two matrices, then to find the derivative of a product, we can apply the product rule from calculus. You should probably prove that the product rule actually works here, but assuming that it does, we find that dx is... Now this is supposed to be equal to this product and sum. So we can find that. And since they're supposed to be equal, we can put them equal to each other. And notice we have this product of matrices appearing on both sides of the equation. So we can drop that product out, leaving us with... which tells us something about c1 prime and c2 prime. Well remember this corresponds to a system of linear equations. So let's write down what that system actually is. We can solve it. Here we might notice that we have an e to the 4t, c2, prime of t in both equations. So if we multiply the first equation by minus one and add, then solve for c1 prime. And since we have c1 prime, we can integrate to find c1 itself. Similarly, we can take our two equations, and if we multiply our top equation by two and add, we'll eliminate the c1t prime. Solving for c2 prime of t gives us... and since we know c2 prime, we can integrate to find c2, which tells us what c2 of t is. And so that gives us our solution.