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Studying for a test? Prepare with these 3 lessons on Second order linear equations.
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Let's do another example of solving a nonhomogeneous linear differential equation with a constant coefficient. And the left-hand side is going to be the same one that we've been doing. The second derivative of y minus 3 times the first derivative minus 4 times y is equal to-- and now instead of having an exponential function or a trigonometric functional, we'll just have a simple-- well, it just looks an x squared term, but it's a polynomial. Right? And you know how to solve the general solution of the homogeneous equation if this were 0. So we're going to focus just now on the particular solution, then we can later add that to the general solution of a nonhomogeneous equation, to get the solution. So what's a good guess for a particular solution? Well, when we had exponentials, we guessed that our solution would be an exponential. When we had trigonometric functions, we guessed that our solution would be trigonomretric. So since we have a polynomial here that makes this differential equation nonhomogeneous, let's guess that a particular solution is a polynomial. And that makes sense. If you take a second-degree polynomial, take its derivatives and add and subtract, you should hopefully get another second-degree polynomial. So let's guess that it is Ax squared plus Bx plus C. And what would be a second derivative? Well a second derivative would be 2Ax plus B. Sorry, this is the first derivitive. The second derivative would be 2A. And now we can substitute back into the original equation. We get the second derivitive, 2A minus 3 times the first derivitive. So minus 3A-- oh no, sorry. Minus 3 times this. So minus 6Ax minus 3B minus 4 times the function itself. So minus 4Ax squared minus 4Bx minus 4C. That's just 4 times all of that. That's going to equal 4x squared. And I'll just group our x squared, our x and our constant terms, and then we could try to solve for the coeficients. So let's see. I have one x squared term here. So it's minus 4Ax squared. And then what are my x terms? I have minus 6Ax minus 4Bx. So then say plus minus 6A minus 4B times x. I just added the coefficients. And then finally we get our constant terms. 2A minus 3B minus 4C. And all of that will equal 4x squared. Now how do we solve for A, B, and C? Well, whatever the x squared coefficients add up on this side, it should equal 4. Whatever the x coefficients add up on this side, it should be equal to 0, right? Because you can view this as plus 0x, right? And then you could say plus 0 constant as well. So the constants should also add up to 0. So let's do that. So first let's do the x squared term. So minus 4A should be equal to 4. And then that tells us that A is equal to minus 1. Fair enough. Now the x terms. Minus 6A, minus 4B, that should be equal to 0. Right? So let's write that down. We know what A is, so let's substitute. So minus 6 times A, so minus 6 times minus 1. So that's 6 minus 4B is equal to 0. So we get 4B-- I'm just putting 4B on this side and then switching. 4B is equal to 6. And B is equal to-- 6 divided by 4 is 3/2. And then finally the constant term should also equal 0, so let's solve for those. 2 times A, that's minus 2. Minus 3 times B. Well, that's minus 3 times this. So minus 9/2 minus 4C is equal to 0. So let's see. I don't want to make a careless mistake. So this is minus 4 minus 9/2, right? That's minus 4/2 minus 9/2-- and we could take the 4C and put it on that side-- it's equal to 4C. What's minus 4 minus 9? That's minus 13/2. Minus 13/2 is equal to 4C. 4C, divide both sides by 4, and then you get C is equal to minus 13/8. And I think I haven't made a careless mistake. So if I haven't, then our particular solution, we now know. Well, let me write the whole solution. So. And this is a nice stretch of horizontal real estate. So let's write our solution. Our solution is going to be equal to the particular solution, which is Ax squared, so that's minus 1x squared. Ax squared plus Bx plus 3/2x plus C minus 13/8. So this is the particular solution. We solved for A, B, and C. We determined the undetermined coefficient. And now if we want the general solution, we add to that the general solution of the homogeneous equation. What was that? y prime minus 3y prime minus 4y is equal to 0. And we've solved this multiple times. We know that the general solution to the homogeneous equation is C1e to the 4x plus C2e to the minus x, right? You just take the characteristic equation r squared minus 3r minus 4. What did you get? You get r minus 4 times r plus 1, and then that's how you get minus 1 and 4. Anyway. So if this is the general solution to the homogeneous equation, this a particular solution to the nonhomogeneous equation. The general solution to the nonhomogeneous equation is going to be the sum of the two. So let's add that. So plus C1e to the 4x plus C2e to the minus x. So there you. I don't think that was too painful. The most painful part was just making sure that you don't make a careless mistake with the algebra. But using a fairly straightforward, really algebraic technique, we were able to get a fairly fancy solution to this second order linear nonhomogeneous differential equation with constant coefficients. See you in the next video.