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Current time:0:00Total duration:10:11

Video transcript

we're now ready to solve non-homogeneous second-order linear differential equations with constant coefficients so what is all that mean well it means a an equation that looks like this a times the second derivative plus B times the first derivative plus C times the function is equal to G of X before I show you an exact actual example I want to show you something interesting that that the general solution of this non-homogeneous equation is actually the general solution of the homogenous equation plus a particular solution and I'll explain what that means in a second so let's say that H is a solution of the homogenous equation H is homogeneous and that worked out well because H for homogeneous H is solution for homogeneous solution for homogeneous there should be some shorthand notation for homogeneous so what is what what does that mean that means that a times the second derivative of H plus B times H prime plus C times H is equal to 0 that's what I mean when I say that H is a solution and actually let's just say that H is the general solution for the homogenous equation and we know how to solve that take the characteristic equation depending on how many roots it has and whether they're real or complex you can figure out a general equation a general solution and then if you have initial conditions you can substitute them and get the values of the constants fair enough now let's say that I were to say that G is a solution let's say it's 8 we'll know how to use G up here let's say what am i well I don't like using vowels let's say J let's say J is a particular solution to this differential equation so what does that mean that means that a times J prime prime plus B times J prime plus C times J is equal to G of X right so that means or we're just defining J of X to be solution to be a particular solution particular solution now what I want to show you is that J of X plus h of X is also going to be a solution to this original equation and that it's it's the general solution for this non-homogeneous equation and and before I just you know do it mathematically what's the intuition well when you substitute H here you get 0 when you substitute J here you get G of X so when you add them together you're going to get 0 plus G of X here so you're gonna get G of X here and I'll show you that right now so let's say so let's say I wanted to substitute H plus J here so what is and I'll do it in a different color a so the second derivative of that of that the sum of those two functions is going to be the second derivative of both of them summed up plus B times the first derivative of the sum plus C times the sum of the functions and my goal is to show that this is equal to G of X so what does this simplify to well if we take all the H terms we get a H prime prime plus B H prime plus C H Plus let's fill the J terms a J prime prime plus B J prime plus C J well by definition of how we defined H and J what is this equal to well we said that H is is a solution for the homogenous equation or that this expression is equal to 0 so that equals 0 and by our definition for J what is this equal when we said J is a particular solution for the non-homogeneous equation or that this expression is equal to G of X so this is equal to G of X so when you substitute h plus J into this differential equation on the left hand side on the right hand side true enough you get G of X so we've just shown that if you define H and J this way that the function I don't know we'll call it K of X is equal to H of X plus J of X I'm running out of space that is the general solution I haven't proven that it is the most general solution but I think you have the intuition right because the general solution of the homogenous one that was the general solution the most general solution and now we're adding a particular solution that gets you the G of X on the right hand side that might be very confusing to you so let's actually try to do it with some real numbers and I think it'll make a lot more sense so let's say we have the differential equations and I'm going to teach you a technique now for figuring out that G or that J in the particulars in that last example so how do you figure out that particular solution so let's say i have the differential equation the second derivative of y minus 3 times the first derivative minus 4 times y is equal to 3 e to the 2x so the first step is we want the hum we want the general solution of the homogenous equation so first we can get and in that example I just did that would have been our H of X so we want the solution of Y prime prime minus 3 y prime minus 4y is equal to 0 take the characteristic equation 4 is equal to 0 our what is this R minus 4 times R plus 1 is equal to 0 two roots are could be 4 or negative 1 and so our general solution I will call that H let's call that Y general y sub G so our general solution is equal to we've done this many times C 1 e to the 4x plus c2e to the minus 1 X or minus fair enough so we solved the homogeneous equation so how do we get a in that last example a J of X that'll give us a particular solution so on the right hand side we get this well here we just have to think a little bit and this this method is called the method of undetermined coefficients and you have to say well if I want something some function when I take its second derivative and add that or subtracted some multiple of its first derivative minus some multiple of the function I get e to the 2x that function in its derivatives in the second derivatives must be something of the form something times e to the 2x so essentially we take a guess we say well what what does it look like when we take the derivatives in a and and the various derivatives and the functions and we multiply multiples of it plus each other and all of that then we get e to the 2x or some multiple of 8 is 2x well a good guess could just be that J which is our well I'll call it Y particular our particular solution here could be that and particular solution I'm using a little different than the particular solution when we we had initial conditions here we can view this as a particular solution a solution that gives us this on the right-hand side so let's say that the one I pick is some constant a times e to the 2x if that's my guess then the derivative of that is equal to 2ei e to the 2x and the second derivative of that of my particular solution is equal to 4ei e to the 2x and now I can substitute in here and let's see if I can solve for a and then I'll have my particular solution so the second derivative that's this so I get 4a e to the 2x minus 3 times the first derivative so minus 3 times this so that's minus 6a e to the 2x minus 4 times the function so minus 4a e to the 2x and all that is going to be equal to 3 e to the 2x well we know e to the 2x doesn't equal zero so we can divide both sides by that just factor it out really get rid of all of these the 2x and the left hand side we have 4a and a minus 4a well those cancel out and then lo and behold we have minus 6a is equal to 3 divide both sides by 6 you get a is equal to minus 1/2 so there we have our particular solution it is equal to minus 1/2 e to the 2x and now I like I just showed you before I cleared the screen our general solution of this non-homogeneous equation is going to be our particular solution plus the general solution to the homogenous equation so our we could call this the most general solution or I'll just call it Y it is our general solution c1 e to the 4x plus oops plus c2 e to the minus X plus our particular solution we find found so that's minus minus 1/2 e to the 2x pretty neat anyway well I'll do a couple more examples of this I think it'll get the hang of it in the next example we'll do something other than an e to the 2x or an e function here we'll try to do stuff with polynomials and and trig functions as well I'll see you in the next video