If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

Undetermined coefficients 2

Another example using undetermined coefficients. Created by Sal Khan.

Want to join the conversation?

  • female robot ada style avatar for user albert
    at , it shouldn't be -34 instead of +34?
    (64 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user cbarajasjr
    Second order DEs require 2 parameters which is why, I think, that we need A and B coefficients, but in "guessing" our function y=Asinx+Bcosx, why not choose y=Asinx+Bsinx or y=Acosx+Bcosx?
    (11 votes)
    Default Khan Academy avatar avatar for user
    • blobby green style avatar for user Jesse.T.Robinson
      Since the equation on the right was 2sinx, he knew that when he plugged it into the differential equation that he would get some sort of cosine from the first derivative and some sort of sine from the second derivative. This is why he wants to make his guess contain both! So he chooses y = Asinx + Bcosx.

      If he had chosen y = Asinx + Bsinx, this could be rewritten as y = (A+B)sinx. This is just y = Csinx where C = A + B (just numbers), so again we're missing a term if we try to use just this. A similar thing happens if you try y = Acosx + Bsinx.

      Let me show you more explicitly what I mean. Try y = Asinx. Then y' = Acosx, and y'' = -Asinx. Plug these into the equation y'' - 3y' - 4y = 2sinx to get

      (-Asinx) - 3(Acosx) - 4(Asinx) = 2sinx
      (-A - 4A)sinx - 3Acosx = 2sinx
      -5Asinx - 3Acosx = 2sinx

      So, matching coefficients,

      -5A = 2
      -3A = 0 (since there is no cosine term, the coefficient is 0)

      This tells us that A = -2/5 but also A = 0, which is not possible! So we do need some sort of cosine term in our guess, and choosing to use y = Asinx + Bcosx works.

      For comparison, check out the previous video where the on the right hand side of equation there was a 3e^(2x). Notice how he only needs to use y = Ae^(2x) because this function only changes numerical coefficients when taking derivatives. The derivative of 2sinx was a different story.

      I hope this helps answer your question!
      (82 votes)
  • duskpin sapling style avatar for user Justo
    Why he takes Asinx+Bcosx and not another function?
    (4 votes)
    Default Khan Academy avatar avatar for user
  • mr pants teal style avatar for user Tristen Wentling
    In this video there is a sine function, then next is a polynomial, that's all good and I pretty well got the hang of those, but how do you chose the particular solution for tangent for example, or secant?
    (5 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Shams Sheenwari
    sir i m confused that how we can select the Y=Asinx +Bcosx , is there any clue that based on what we can select this term and we can interprete that for other differential equations

    (5 votes)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Mohammad Abdollah
    Thank you very much for these course they really help and simplify lot of complecated formulas, however sometimes the solution are given as guesses and not every one can guess and get the right answer, for example for this video your guess about yp= Acosx+Bsinx we can't guess every problem there must be a way to find this particular solution. please advise I read a way called finding the basis it not clear to me if you can please explain how can we find thoses basis. Thanks
    (3 votes)
    Default Khan Academy avatar avatar for user
  • leafers ultimate style avatar for user prasadshivshankar5
    I'm not sure, but shouldn't the third equation be 4*Asin(x) *+ 4*Bcos(x)? Since the coefficient on the y term in the original equation is a *negative four, and he calculates y" to be -Asin(x)-Bcos(x), the negatives should cancel out...right? Thanks
    (2 votes)
    Default Khan Academy avatar avatar for user
  • male robot hal style avatar for user Saad Saeed
    Couldn't you have used Cramer's rule to solve for A and B?
    (2 votes)
    Default Khan Academy avatar avatar for user
  • leafers seed style avatar for user Zulfidin Hojaev
    at when I put A= -5/17 into (-5A+3B=2) i got B=7/51 , which is different than what sal found
    (1 vote)
    Default Khan Academy avatar avatar for user
  • blobby green style avatar for user Tarik Muhovic
    I got f''(t)-6f'(t)+9f(t)=20sin(2t)-48cos(2t).

    I got it down to:
    5A+12B = 20
    -12A+5B = -48

    How would i go about solving A and B? The steps at confused me a little.
    (1 vote)
    Default Khan Academy avatar avatar for user

Video transcript

Let's do some more nonhomogeneous equations. So let's take the same problem, but we'll change the right-hand side. Because I think you know how to solve the-- essentially, the homogeneous version. So the same problem as we did in the last video. The second derivative of y minus 3 times the first derivative y minus 4 times the function. And now in the last example, the nonhomogeneous part was 3e to the 2x. But we're tired of dealing with exponent functions, so let's make it a trigonometric function. So let's say it equals 2 sin of x. So the first step you do is what we've been doing. You essentially solve the homogeneous equation. So this left-hand side is equal to 0. You do that by getting the characteristic equation r squared minus 3r minus 4 is equal to 0. You get the solutions, r is equal equal to 4, r is equal to minus 1, and then you get that general solution. We did this in the last video. You get the general solution of the homogeneous. Maybe we'll call this the homogeneous solution. y homogeneous. We've got the C1 e to the 4x plus C2e to the minus x. And that's all and good, but in order to get the general solution of this nonhomogeneous equation, I have to take the solution of the nonhomogeneous equation, if this were equal to 0, and then add that to a particular solution that satisfies this equation. That satisfies-- when you take the second derivative minus 3 times the first minus 4 times the function, I actually get 2 sin of x. And here once again we'll use undetermined coefficients. And undetermined coefficients, just think to yourself. What function, when I take its second and first derivatives and add and subtract multiples of them to each other, will I get sine of x? Well, two functions end up with sine of x when you take the first and second derivatives. And the sine and cosine of x. So it's a good guess. And that's really what you're doing it the method of undetermined coefficients. You take a guess of a particular solution and then you solve for the undetermined coefficients. So let's say that our guess is y is equal to-- I don't know, some coefficient times sine of x. And if this was sine of 2x, I'd put A times sine of 2x here. Just because I still want-- no matter what happens here-- the sine of 2x's or maybe cosine of 2x's to still exist. If this was a sine of 2x, there's nothing I could do to a sine of x, or nothing at least trivial that I could do to the sine of x. It would end up with a sine of 2x. So whatever's here, I want here. Plus B, some undetermined coefficient times cosine of x. And once again, this was sine of 2x. I'd want a cosine of 2x here. So let's figure out its first and second derivatives. So the first derivative of this y prime is equal to A cosine of x. Cosine derivative is minus sine, so minus B sine of x. And then the second derivitive-- I'll write down here. The second derivative is equal to what? Derivative of cosine is minus sine, so minus A sine of x minus B cosine of x. I think you're starting to see that the hardest thing in most differential equations problems is not making careless mistakes. It's a lot of algebra and a lot of fairly basic calculus. And the real trick is to not make careless mistakes. Every time I say that, I tend to make one. So I'm going to focus extra right now. So anyway, let's take these and substitute them back into this nonhomogeneous equation. Let's see if I can solve for A and B. So the second derivative is that. Let me just rewrite it, just so that you see what I'm doing. So I'm going to take the second derivative, y prime prime, so that's minus A sine of x minus B cosine of x. I'm going to add minus 3 times the first derivative to that. And I'm going to write the sines under the sines and the cosines under the consines. So minus 3 times this. So the sine is, let's see. It's plus 3B sine of x minus 3 times this. So minus 3A cosine of x. And then minus 4 times our original function. So minus 4A sine of x. Right? Minus 4 times that. Minus 4 times this. Minus 4B cosine of x. And when I take the sum of all of those-- that's essentially the left-hand side to this equation-- when I take the sum of all of that, that is equal to 2 sine of x. I could have written them out in a line, but it would have just been more confusing. And now this makes it easy to add up the sine of x's and the cosine of x's. So if I add up all the coefficients on the sine of x, I get minus A plus 3B minus 4A. So that looks like minus 5A plus 3B sine of x plus-- and now what are the coefficients here? I have minus B and then I have another minus 4B, so minus 5B and then minus 3A. So minus 3A minus 5B cosine of x. The cosine of x should go right here. So anyway, how do I solve for A and B? Well, I have the minus 5A 3B is equal to whatever coefficients in front of sine of x here. So minus 5A plus 3B must be equal to 2. And then minus 3A minus 5B is the coefficient on cosine of x, although I kind of squeezed in the cosine of x here, right? So this must be equal to whatever the coefficient on cosine of x is on the right-hand side. Well the coefficient of cosine of x on the right-hand side is 0. So that sets up a system of two unknowns with two equations. A linear system. So we get minus 5A plus 3B is equal to 2. And we get minus 3A minus 5B is equal to 0. And let's see if I can simplify this a little bit. Let's see. This is a system of two unknowns, two equations. If I multiply the top equation by 5 1/3s, right? Actually, let me multiply the top equation by 5 1/3s. I get minus 25/3 A plus 5B is equal to 5 1/3s times this. 5 1/3s times 2 is 10 1/3s. And the bottom equation is minus 3A minus 5B is equal to 0. Let's add the two equations. I get 10 1/3s is equal to-- these cancel out. That's minus 25/3 minus 9/3 A is equal to 10 1/3s. This is getting a little bit messier than I like, but we'll soldier on. So minus 25 minus 9. What's minus 25 minus 9? So that is 34. So we get 34 over 3A is equal to 10/3. We can multiply both sides by 3. Divide both sides by 34. A is equal to 10/34, which is equal to 5/17. Nice ugly number. 5/17 and now we can solve for B. So let's see. Minus 3 times A minus 3 times A. 5/17 minus 5B is equal to 0. So that's what? Minus 15/17 is equal to plus 5B. I just took this and put it on the right-hand side. And then divide both sides by 5. Oh, you know what? I realized I made a careless mistake here. Minus 25 minus 9. That's the minus 34 over 3. so minus 34A is equal to 10. A is equal to minus 10/34 or minus 5/17. So minus 3 times minus 5/17. So 5/17 is equal to plus 5B, right? And then we get B is equal to 3/17. That was hairy. And notice, the hard part was not losing your negative sines. But anyway, we now have our particular solution to this. let me try to write in a non-nauseating color, although I think I picked a nauseating one. The particular solution is A minus 5/17 sine of x-- right? That was a coefficient on sine of x-- plus B plus 3/17 times cosine of x. And if we look at our original problem, the general solution out of this nonhomogeneous equation would be this-- which is the general solution to the homogeneous equation, which we've done many videos on-- plus now our particular solution that we solved using the method of undetermined coefficient. So if you just take that and add it to that, you're done. And I am out of time. See you in the next video.