 So there's one more situation where we have to modify our approach to solving differential equations using our linear differential operator So remember if we could write a linear differential equation in the form l applied to y equals zero Where l is a linear differential operator then the solutions will be linear combinations of C e to power rx and e to power ax c1 cosine bx plus c2 sine bx where r and a plus or minus bi are roots of the characteristic polynomial And again the assumption is that we can somehow find these roots algebraically numerically or through wild and wishful thinking But sometimes that isn't enough. What about an equation like d2 y plus 6 dy plus 9 y equals zero So let's rewrite this in operator form Our characteristic polynomial is going to be d2 plus 6d plus 9 We'll find the roots by setting the characteristic polynomial equal to zero and solving and so we get roots d equals minus 3 and d equals minus 3 and So this gives us a general solution y equals c1 e to power minus 3x And while this is a general solution, this doesn't actually include all of the solutions Remember that since this is a second order differential equation We're supposed to have two unknown constants and we only have one We must be missing a whole set of solutions. So where are they? Since y equals c1 e to power minus 3x is one solution We'll suppose that there is another solution But what? A common approach is to assume that our constant c1 is actually a function c1 of x This is known as variation of parameters and we'll have a lot more to say about it later But the first question that comes to mind is why do we get to make this assumption? And the answer is ah, let's see what happens This goes back to an important idea in mathematics and in life. It's easier to obtain forgiveness than permission If it works great if it doesn't work. Well, then you'll have some explaining to do So we'll have a lot more to say about variation of parameters later But in this particular case if the differential operator has r as a root of multiplicity k The corresponding solutions will be our first solution some constant times e to power rx To get the other solutions will multiply by powers of x So a second solution c2 x e to power rx a third solution c3 x squared e to power rx and so on until we get enough Solutions to account for the multiplicity of the root So back to our original problem Rewriting our differential equation using the differential operator And so our characteristic polynomial is d squared plus 6d plus 9, which is the same as d plus 3 squared This has root minus 3 with multiplicity 2 and So the general solution was going to be c1 e to power minus 3x, which we get for the root of minus 3 Since our roots have multiplicity 2 we get a second solution x e to power minus 3x and so our solutions are going to be linear combinations of these two and as Required a second order linear differential equation has solution with two unknown constants