 So we can rewrite any linear differential equation in operator form Ly equals h of t using a linear differential operator L Where L can be viewed as a characteristic polynomial in D the differential operator So all of this is very nice and it allows us to rewrite a differential equation in a much simpler form But does this do us any good? In other words, how can this help us solve differential equations? So suppose L is a linear differential operator and y1 and y2 are two of the solutions to Ly equals 0 Then for any constant c1 and c2 What do we get when we apply L to the linear combination c1 y1 plus c2 y2? Remember the linearity property of our linear differential operator says that if we apply L to a sum It's the sum of L applied to the individual sum ads So this is L applied to c1 y1 plus L applied to c2 y2 But again the linearity of L means that L applied to c1 y1 Since c1 is a constant we can remove that constant factor to the front So this is going to be c1 L applied to y1 And similarly L applied to c2 y2. Well, that's just going to be c2 times L applied to y2 But wait, there's more since y1 and y2 are solutions to our differential equation Ly equals 0. This means that L applied to y1 is 0 and L applied to y2 is also 0 And so this expression simplifies to 0 and consequently Any linear combination of solutions to Ly equals 0 is a solution to Ly equals 0 In other words linear combinations of solutions will also give you a solution So that's amazing. That's wonderful. That's extremely useful. Once we have a solution we can find more Except how do we find solutions? Well, let's consider a few simple cases So let's consider the simplest characteristic polynomial d minus a where a is a real number This characteristic polynomial corresponds to the differential equation d minus a applied to y equals 0 And let's rewrite this in a more familiar form Remember, this is the differential operator So this is really the differential operator applied to y minus a times y Or writing this as a differential equation dy over dx minus a y equals 0 And this is a separable differential equation. So we'll solve it And so our differential equation has solution y equals c e to power a x And so now we know what to do if our characteristic polynomial is linear Now admittedly this isn't very exciting because our linear characteristic polynomials Correspond to separable differential equations and we already know how to solve those so it doesn't seem like we've really gained anything So now let's consider the characteristic polynomial l and suppose r is a root Then l can be written in the form l star applied to t minus r Where l star is another characteristic polynomial So let's consider what happens if we try to apply l to our solution c e to power r x So l applied to c e to the r x well That's really l star applied to d minus r applied to c e to power r x So if I apply d minus r to c e to the r x that's really d applied to c e to the r x minus r times C e to the r x Remember this is just the derivative of this function, which is going to be and so this becomes and That means l star is going to be applied to zero But since l star is a linear differential operator It consists of a whole bunch of derivatives Well, if I take the derivative of zero and then add a constant time zero I get zero And so this means that c e to power r x is a solution to our original differential equation And this gives us an amazingly powerful method of solving higher-order differential equations Let l be the linear differential operator for differential equation if r is a root of the corresponding characteristic polynomial c e to power r x is a solution to the differential equation And since linear combinations of solutions will also be solutions This means we can find a general solution which will look like a linear combination of terms with factor e to power r x So for example, let's find two different families of solutions to the differential equation y double prime minus 3 y prime equals minus 2 y So first we'll rewrite the equation in operator form And this means we need to get all of our y terms over onto the left-hand side so we'll rewrite This second derivative is d2 applied to y This first derivative is d applied to y and 2 y is just 2 y Not factoring, but rewriting in function notation This is the same as d2 minus 3d plus 2 applied to y And so our linear differential operator can be written as d squared minus 3d plus 2 The roots of the characteristic polynomial occur where the polynomial is equal to 0 So we'll set that polynomial equal to 0 and solve And so the roots are d equals 2 and d equals 1 So we can summarize our results so far The characteristic polynomials d squared minus 3d plus 2 with roots d equals 2 and d equals 1 Our theorem says that each of these roots corresponds to a solution of the form e to power r x And so our solutions to the differential equation are going to be y equals c1 e to power 2 x and y equals c2 e to power x And a general solution will be a linear combination of these And a useful check is to remember that you should have as many Undetermined constants as the order of the differential equation This is a second-order differential equation. We should have two undetermined constants c1 and c2