Okay, so for this differential equation, we need to find solutions of the form. Why is equal to Exeter? Are so then why Prime is going to be our X to the R minus one. And then why double time is going to be equal to our times? Ar minus one times, actually. Ar minus two. So now substituting these into this differential equation Here we have two x squared. Our times are minus one times times X to the R minus two plus five x Times are times actually ar minus one. It's plus exit E r is equal to zero. Um, we can simplify this, uh, so this x squared and this actually ar minus who becomes extra the are and then the two and the arse. Times ar minus one. I'm going to turn that into two R squared minus to our Okay. Then move that. You know that a little here and then similarly, here this x and X R minus one becomes our or extra that are so we just get five are times actually are plus X to the are is zero. So now we can see. Um, we can, um, back throughout an extra are from all three terms. So we have. Actually, there are. And then what's left over is to our square minus two are plus five r plus one is equal to zero. Since XLR is not gonna be equal to zero for X greater than zero. Now we just need to set this part of the creation equals zero. So I get to r squared and then minus two r plus five are that becomes plus three are plus one is equal to zero. Okay, so, um, here, since we only have one here, but we have to hear, um, we can factor into to our and then that's just gonna be one here and one here, since everything's positive will just have positive there. Okay, so this does work out. So too are, um is R squared then to our and then our is three r, and then one is one. Okay, so our two solutions are going to be negative 1/2 and negative one. So we're going to have X to the negative 1/2 and then X to the negative one. Eyes are to Linnean independent solution. So our general solution Why? See, that's gonna be c one c one x to the negative 1/2 um, negative 1/2 plus c two x the negative one.