 In this video, we provide the solution to question number two for practice exam number three for math 1220 in which case we're given a differential equation x squared times y prime is equal to y. We also have some initial value y equals negative one when x equals one, and we also have the general solution to this differential equation, which we don't need to then compute it. We already know it. We want to find a particular solution here, which be aware if we already have the general solution and we have the initial value, turns out we don't need the differential equation whatsoever. It comes down to plugging these numbers into the general solution and finding the unknown value see there. So if y is equal to negative one, we're going to get negative one equals c times e to the negative one over x here, which is going to be one. If you simplify the right hand side, you're going to get c to the negative c times e to the negative one, which of course is c over e. This is equal to negative one. If you times both sides by e, you'll get that c equals negative e, and therefore that's the coefficient we need there. If you plug in a negative e for the c value, well, since you have two ease there, I can actually put the, I can actually, since this is an exponent of one there, you can add those together. So I really could write this as y equals negative e to the one minus one over x power, like so, and for which if you search among the options, you see that's exactly option c, which then is the correct answer.