2nd order linear homogeneous differential equations 3

in the last video we had this second-order linear homogeneous differential equation and we just tried out the solution Y is equal to e to the RX and we got we figured out that if you try that out then it works for particular ARS and those ARS we figured out the last one were minus 2 and minus 3 but it came out of factoring this characteristic equation and watch the last video if you forgot how we got that characteristic equation and we ended up with this general solution for this differential equation and you could try it out if you don't believe me that it works but what if we have we don't want the general solution we want to find the particular solution well then we need initial condition so let's let's use let's do this differential equation with some initial condition so let's say the initial conditions are let's say let me scroll this down we have the solution that we figured out in the last video and let me rewrite the differential equation so it was the second derivative the second derivative plus five times the first derivative plus 6 times the function is equal to 0 and the initial conditions were given is that Y of 0 is equal to 2 and the first derivative at 0 or Y Prime at 0 is equal to 3 so they're giving us what does y equal at the point 0 and what is the slope and this at 0 at X is equal to 0 and the slope is 3 so how do we use these to solve for c1 and c2 well let's just use the first initial condition Y of 0 is equal to 2 so Y of 0 so Y of 0 is equal to 2 which is equal to essentially just substitute 0 and into this equation so it's c1 times e to the minus 2 times 0 what's either that's essentially e to the 0 right so that's just 1 so let's see 1 times 1 which is just c1 plus c2 times e to the minus 3 times 0 this is e to the 0 so it's just 1 so plus c2 so the first equation we get when we substitute our first initial condition is essentially c1 plus c2 is equal to 2 now let's apply our second initial condition that tells us the slope at X is equal to 0 so Y prime is 0 so this is our general solution let's take its derivative and then we can use this so Y prime of X Y prime of X is equal to what the derivative of this is equal to minus 2 minus 2 c1 times e to the minus 2x and what's the derivative of this it's minus 3 c2 times e to the minus 3x and now we can use our initial condition y Prime at 0 so when X is equal to 0 what's the right-hand side equal it's minus 2 times c1 and then e to the minus 0 e to the 0 so that's just 1 minus 3 c2 and then once again X is 0 so e to the minus 3 times 0 that's just 1 so it's just 1 times minus 3 c2 and it tells us that when X is equal to 0 what is this whole derivative equal well that equals 3 right Y prime of 0 is equal to 3 so now we go back into your first year of algebra we have two equations two linear equations with two unknowns and we could solve let me write them in a form that you're probably more used to so the first one is c1 plus c2 is equal to 2 and the second one is minus 2 c1 minus 3 C 2 is equal to 3 is equal to 3 so what can we do let's multiply this top equation by 2 there's a ton of ways to solve this but if you multiply the top equation times 2 you'll get and I'll do this in a different color just so that's changed I'm just multiplying the top one by 2 you get 2 C 1 plus 2 C 2 is equal to 4 and now we can add these two equations let's see minus 2 C 1 plus 2 of those cancel out so minus 3 plus 2 you get minus C 2 is equal to seven or we could say that C 2 is equal to minus seven and now we can substitute back in here we have C 1 plus C 2 C 2 is minus 7 so minus 7 is equal to 9 or we know that C oh sorry no I'm already confusing myself my brain was getting ahead of myself C 1 plus 2 C 2 that's minus 7 is equal to 2 right I'm just substituting back to this differential equation let's sign into this equation not a differential is just a simple linear equation and then we get C 1 is equal to 9 and now we have our particular solution to the differential equation so this was our general solution we can just substitute our C ones and our C 2's back in we have our particular solution for those initial conditions and I think that warrants a different color so our particular solution is y of X is equal to C 1 which we figured out is 9 e to the minus 2 X plus C 2 well C 2 is minus 7 minus 7 e to the minus 3 X that is the particular solution to our original differential equation and it might be a good exercise for you to actually test it out this particular solution to this differential equation I'll do another example in the next video I'll see you soon