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2nd order linear homogeneous differential equations 4

Video transcript

let's solve another second-order linear homogeneous differential equation and this one well I won't give you the details before I actually write it down so the differential equation is four times the second derivative of Y with respect to X minus eight times the first derivative plus three times the function times y is equal to zero and we have our initial conditions Y of zero is equal to two and we have Y prime of zero is equal to one-half now I could go into the whole thing you know Y is equal to e to the rx is the solution substitute it in then factor out e to the rx and have the characteristic equation and and if you if you want to see all of that over again you might want to watch the previous video just to see where that characteristic equation comes from but in this video I'm just showing you I'm just going to show you literally how how quickly you can do these type of problems mechanically so if this is our original differential equation the characteristic equation is going to be and I'll do this in a different color for R squared minus 8 R plus 3 R is equal to 0 and watch the previous video if you don't know where this characteristic equation comes from but if you want to do these problems really quick you just substitute the second derivatives with an R squared the first derivatives with an R and then the function with with oh sorry no no this this must be a constant so and then and then the coefficient on the original function is just the constant right I think you see what I did right second derivative R squared first derivative R no derivative you could say that's R to the 0 or just 1 but this is our characteristic equation and now we can just figure out its roots this is not a trivial one for me to factor so if it's not trivial you can use the quadratic equation so we could say the solution of this is R is equal to negative B well negative B be B is negative 8 so it's positive 8 8 plus or minus the square root of b squared so that's 64 minus 4 times a which is 4 times C which is 3 all of that over 2a 2 times 4 is 8 that equals 8 plus or minus square root of 64 - what's 16 what's 16 times 3 times minus 48 all of that over 8 what's 64 minus 48 let's see 12 it's 16 right right 10 plus 48 is 58 then another so it's 16 so we have R is equal to 8 plus or minus the square root of 16 over 8 is equal to 8 plus or minus 4 over 8 that equals 1 plus or minus 1/2 so the two solutions of this characteristic equation and ignore that let me scratch that out in black so that you know that that's not like a 30 or something the two solutions of this characteristic equation are R is equal to well 1 plus 1/2 is equal to three-halves and r is equal to 1 minus 1/2 is equal to 1/2 so we know our two R's and we know that from previous experience in the last video that you know y is equal to c times e to the RX is a solution so the general solution of this differential equation is y is equal to c1 times e let's use our first our e to the 3 halves X plus c2 times e to the 1/2 X this differential equations problem was literally just a problem in using the quadratic equation and once you figure out the RS you have your general solution and now we just have to use our initial conditions so to know the initial conditions we needed a y of X and we need a y prime of X so let's just do that right now so what's Y prime Y prime of our general solution is equal to three-halves times c1 e to the three halves X plus derivative the inside one half times c2 e to the one-half X and now let's use our actual initial conditions I don't want to lose them let me write them down here so I can scroll down so we know that Y of 0 is equal to 2 and Y prime of 0 is equal to 1/2 those are our initial conditions so let's use that information so Y of 0 so Y of 0 what what happens when you substitute X is equal to 0 here you get c1 times e to the 0 essentially so that's just 1 plus c2 well that's just e to the 0 again because X is 0 is equal to so this is when X is equal to 0 what is y y is equal to 2 y of 0 is equal to 2 and then let's use the second equation so when y when we substitute X is equal to 0 in the derivative so when x is 0 we get three-halves c1 this goes to 1 again plus 1/2 1/2 C 2 this is 1 again e to the 1/2 times 0 is e to the 0 which is 1 is equal to so when x is 0 for the derivative Y is equal to 1/2 are those derivative is 1/2 at that point or the slope is 1/2 at that point and now we have two equations in two unknowns and we could do a couple we could solve it a ton of ways I think you know how to solve them let's multiply the top equation I don't let's multiply it by three halves and what do we get we get I'll do it in a different color we get three-halves c1 three-halves c1 plus three-halves c2 is equal to what's three-halves times two it's equal to three and now let's subtract well I don't want to confuse you so let's just subtract the bottom from the top so this cancels out what's one half minus three halves one half minus one and a half well that's just minus one right so minus c2 is equal to what's one half minus three that's minus two and a half or minus five halves minus five halves and so we get c2 is equal to five halves and we can substitute back in this top equation c1 plus five halves is equal to 2 or c1 is equal to 2 which is the same thing as four halves minus five halves which is equal to minus one half and now we can just substitute c1 and c2 back into our general solution and we have solved we have found the particular solution of this differential equation which is y is equal to c1 c1 is minus 1/2 minus 1/2 e to the three halves X three halves X plus c2 c2 is 5 halves plus C 2 which is 5 halves e to the what was this up here e to the one-half X and we are done and it might seem really fancy we're solving a differential equation our solution has e in it and we're taking derivatives and we're doing all sorts of things but really the meat of this problem was solving a quadratic which was our characteristic equation and watch the previous video if you just to see where this you know why this characteristic equation works but it's very easy to come up with the characteristic equation right I think you obviously see that Y prime turns into R squared y prime turns into R and then Y just turns into one essentially so you solve a quadratic and then after doing that you just have take one derivative because after solving the quadratic you immediately have the general solution then you take its derivative use your initial conditions you have a system of linear equations which is Algebra one and then you solve them for the two constants C 1 and C 2 and you end up with your particular solution and that's all there is to it I will see you in the next video