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### Course: Differential equations > Unit 2

Lesson 3: Method of undetermined coefficients# Undetermined coefficients 2

Another example using undetermined coefficients. Created by Sal Khan.

## Want to join the conversation?

- at8:32, it shouldn't be -34 instead of +34?(64 votes)
- Yes. It's a mistake. It should be A = -5/17 and B = 3/17(63 votes)

- 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)
- 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)

- Why he takes Asinx+Bcosx and not another function?(4 votes)
- You want to end up with 2*sin(x) at the end. The derivatives of sin are cos and sin so both must be in your solution but you don't know how many of each, so you have to find an A and B that will make it so after all the steps on the lhs are followed you will be left with 2 sines and zero cosines. Does that make sense?(14 votes)

- 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)
- For this you would have to use another method called variation of parameters, secant and tangent cannot be solved using undetermined coefficients.(4 votes)

- 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

Thanks(5 votes)- It's mostly based on the fact that if y = sin(x), then y' = cos(x) and then y'' = -sin(x). He mostly did this to make it so that the sines cancel, as well as the resultant cosines.(2 votes)

- 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)
- For these problems, there are three basic types: an exponential, a polynomial, and a cosine or sine function. For each of these types there is a standard guess that you can make.(1 vote)

- 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)- You've got the order wrong — at6:04in the video the expressions listed from top to bottom are y'', then -3y', then -4y. So the third equation is -4y = -4(Asin(x) + Bcos(x)), which is what Sal writes.(1 vote)

- Couldn't you have used Cramer's rule to solve for A and B?(2 votes)
- at10:01when I put A= -5/17 into (-5A+3B=2) i got B=7/51 , which is different than what sal found(1 vote)
- 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 at7:28confused me a little.(1 vote)

## 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.