Method of undetermined coefficients
Undetermined Coefficients 3 Another example where the nonhomogeneous part is a polynomial
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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.
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