![]() Divide both sides by 6 and getĪ is equal to minus 1/2. Well we know e to the 2x equalĠ, so we can divide both sides by that. So minus 4Ae to the 2x, andĪll of that is going to be equal to 3e to the 2x. Here, and let's see if I can solve for A, and then I'll have That, of my particular solution, is equal If that's my guess, then theĭerivative of that is equal to 2Ae the to 2x. Here we can view this asĪ particular solution. Our particular solution hereĬould be that- and particular solution I'm using a littleĭifferent than the particular solution when we had Like when we take the various derivatives and the functionsĪnd we multiply multiples of it plus each other? And all of that. Of the form, something times e to the 2x. I want some function where I take a second derivative andĪdd that or subtracted some multiple of its first derivativeĭerivatives and its second derivatives must be something Solution, so on the right-hand side we get this. So we solved the homogeneousĮxample, a j of x that will give us a particular To- and we've done this many times- C1 e to the 4x ![]() Prime prime minus 3y prime minus 4y is equal to 0. And in that example I justĭid, that would have been our h of x. The general solution of the homogeneous equation. Times the first derivative minus 4 times y is equal Particular solution? Let's say I have theĭifferential equation the second derivative of y minus 3 Technique now for figuring out that j in that last example. You, so let's actually try to do it with someĭifferential equations- and I'm going to teach you a On the homogeneous one that was the most general solution,Īnd now we're adding a particular solution that gets Intuition, right? Because the general solution Most general solution, but I think you have the It k of x is equal to h of x plus j of x. So we've just shown that if youĭefine h and j this way, that the function, we'll call J into this differential equation on the left-handĮnough, you get g of x. Solution for the non-homogeneous equation, or J, what does this equal? We said j is a particular The homogeneous equation, or that this expression We defined h and j, what is this equal to? We said that h is a solution for We get Ah prime prime plus Bh prime plus Ch plus, So what is this simplified to? Well if we take all the h terms, The first derivative of the sum plus C times the sum Second derivative of both of them summed up- plus B times Of the sum of those two functions is going to be the And before I just do it mathematically, what's the intuition? Well, when you substitute Solution for this non-homogeneous equation. That j of x plus h of x is also going to be a solution Prime plus B times j prime plus C times j So what does that mean? That means that A times j prime And then if you have initialĬonditions, you can substitute them and get the values Take the characteristic equationĭepending on how many roots it has and whether they're ![]() Is the general solution for this homogeneous equation. That h is a solution- and actually, let's just say that h Second derivative of h plus B times h prime plus C times So what does that mean? That means that A times the And that worked out well,īecause, h for homogeneous. Is a solution of the homogeneous equation. Of the homogeneous equation plus a particular This non-homogeneous equation is actually the general solution Before I show you an actualĮxample, I want to show you something interesting. Plus B times the first derivative plus C times theįunction is equal to g of x. So what does all that mean? Well, it means an equation ![]() Non-homogeneous second-order linear differential equations Nevertheless, finding even one solution doesn't always turn up to be as easy as it may sound. In the example we could have guessed e^(-x) - 1/2 e^(2x), checked that this is in fact a solution, and used that to write all solutions (by adding the same parametrized homogeneous solution). When looking for a particular solution (solution to the original, inhomogeneous equation) it's enough to find one, no matter which one. The part of the video from 17' to 28' also includes a bit more general demonstration of what Sal shows in his video - that such sums actually are solutions - as well as an explanation of the "particular solution" language. You may find a proof that such sums cover all the possible solutions in the 12th lecture of the 18.03 course. ![]() In the inhomogeneous linear case every solution may be expressed as a sum of an arbitrary solution to the inhomogeneous equation plus a solution to the associated homogeneous equation. \) are substituted into Equation 2.2.Differential equations in general have a whole class of solutions, each making the equality true. ![]()
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