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Studying for a test? Prepare with these 3 lessons on Second order linear equations.
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Before we move on past the method of undetermined coefficients, I want to make and interesting and actually a useful point. Let's say that I had the following nonhomogeneous differential equation: the second derivative of y minus 3 times the first derivative minus 4y is equal to-- now this is where gets interesting-- 3e to the 2x plus 2 sine of x plus-- let me make sure that I'm doing the same problems that I've already worked on-- plus 4x squared. So you might say, wow, this is a tremendously complicated problem. I have the 3 types of functions I've been exposed to, I would have so many undetermined coefficients, it would get really unwieldy. And this is where you need to make a simplifying realization. We know the three particular solutions to the following differential equations. We know the solution to the second derivative minus 3 times the first derivative minus 4y. Well, this is this the homogeneous, right? And we know that the solution to the homogeneous equation-- we did this a bunch of times-- is C1e to the 4x plus C2e to the minus x. We know the solution to-- and I'll switch colors, just for a variety-- y prime prime minus 3y prime minus 4y is equal to just this alone: 3e to the 2x. And we saw that that particular solution there, y particular, was minus 1/2 e to the 2x. And we did this using undetermined coefficients. We did that a couple of videos ago. And then let me just write this out a couple of times. We know the solution to this one, as well. This was another particular solution we found. I think it was two videos ago. And we found that the particular solution in this case-- and this was a fairly hairy problem-- was minus 5/17 x plus 3/17. Sorry. The particular solution was minus 5/17 sine of x plus 3/17 cosine of x. And then finally this last polynomial, we could call it. We know the solution when that was just the right-hand side. That was this equation. And there we figured out-- and this was in the last video. We figured out that the particular solution in this case was minus x squared plus 3/2 x minus 13/8. So we know the particular solution when 0's on the right-hand side. We know it when just 3e to the 2x is on the right-hand side. We know it just when 2 sine of x is on the right-hand side. And we know it just when 4x squared is on the right-hand side. First of all, the particular solution to this nonhomogeneous equation, we could just take the sum of the three particular solutions. And that makes sense, right? Because one of the particular solutions, like this one when you put it on the left-hand side, it will just equal this term. This particular solution, when you put it in the left-hand side, will equal this term. And finally, this particular solution, when you put it on the left-hand side, will equal the 4x squared. And then you could add the homogeneous solution to that. You put it in this side and you'll get 0. So it won't change the right-hand side. And then you will have the most general solution because you have these two constants that you can solve for depending on your initial condition. So the solution to this seemingly hairy differential equation is really just the sum of these four solutions. Let me clean up some space because I want everything to be on the board at the same time. So the solution is going to be-- well, I want that to be deleted. I'll do it in baby blue. Is going to be the solution to the homogeneous C1e to the 4x plus C2e to the minus x minus 1/2 e the to 2x. And I'll continue this line. Minus 5/17 sine of x plus 3/17 cosine of x minus x squared plus 3/2 x minus 13/8. And it seems daunting. When you saw this, it probably looked daunting. This solution, if I told you this was a solution and you didn't know how to do undetermined coefficients, you're like, oh, I would never be able to figure out something like that. But the important realization is that you just have to find the particular solutions for each of these terms and then sum them up. And then add them to the general solution for the homogeneous equation, if this was a 0 on the right-hand side. And then you get the general solution for this fairly intimidating-looking second order linear nonhomogeneous differential equation with constant coefficients. See you in the next video, where we'll start learning another method for solving nonhomogeneous equations.