Undetermined coefficients 2 Another example using undetermined coefficients.
Undetermined coefficients 2
- 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 derivitives.
- 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.
- 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
- 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.
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At 5:31, how is the moon large enough to block the sun? Isn't the sun way larger?
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When naming a variable, it is okay to use most letters, but some are reserved, like 'e', which represents the value 2.7831...
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This is great, I finally understand quadratic functions!
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