 Any time you have a method of solving a system of linear equations, you find that a lot of problems can be reduced to solving a system of linear equations. And in differential equations, one of the surprising applications here is to reduce the order of a differential equation. And this comes about as follows. Suppose we have an nth order linear differential equation. We can reduce this to a system of first-order linear differential equations by the substitutions x0 is equal to y, x1 is dy dt, x2 is our second derivative, and so on, all the way up to our n minus first derivative. Now, if we make these substitutions, our original equation becomes which is a first-order differential equation. And the other variables can be expressed in terms of first-order differential equations. For example, let's try to solve this second-order differential equation. So the thing to remember here is that in an nth order differential equation, we'll substitute for the function and the first n minus one derivatives. So x0 is going to be the function itself, x1 is going to be the first derivative, and we don't need any higher derivatives. We do, however, need the derivatives of x0 and of x1. So x0 prime, that's just the derivative of y with respect to t, and x1 prime, well, that's going to be the second derivative of y with respect to t. And so our equation becomes, we'll replace the second derivative with x1 prime. Now, dy dt could be replaced with either x1 or x0 prime. And the general rule here is we only want one derivative per equation. So we already have a derivative x1 prime. So we'll want to replace dy dt with x1. And finally, we could replace y with x0. And so rearranging our equation a little bit. And so using the substitution x0 equals y, x1 equals dy dt, our equation becomes the system where we have the derivatives of x0 and x1. Now, again, the general rule is we only want to have one derivative per equation. So x0 prime is dy dt, but that gives us two derivatives. But since dy dt is x1, then we can write x0 prime is equal to x1. We already have an equation for x1 prime that only involves the single derivative, so we'll use that as our second equation. So we'll rewrite our system of equations in matrix form, and now we can solve it. So our coefficient matrix has eigenvalues lambda equal 1 with eigenvector 1, 1, and lambda equal negative 2 with eigenvector negative 1, 2, and so we get our general solution. And remember x0 is equal to y. So since x0 equals y, this gives us the solution y equals the first components, where we would ordinarily get minus 1 times c2, but we've absorbed the constant into c2.