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## Differential equations

### Unit 2: Lesson 3

Method of undetermined coefficients# Undetermined coefficients 4

Putting it all together! Created by Sal Khan.

## Video transcript

Before we move on past the
method of undetermined coefficients, I want to make and
interesting and actually a useful point. Let's say that I had the
following nonhomogeneous differential equation: the
second derivative of y minus 3 times the first derivative minus
4y is equal to-- now this is where gets interesting--
3e to the 2x plus 2 sine of x plus-- let me
make sure that I'm doing the same problems that I've
already worked on-- plus 4x squared. So you might say,
wow, this is a tremendously complicated problem. I have the 3 types of functions
I've been exposed to, I would have so many
undetermined coefficients, it would get really unwieldy. And this is where you need
to make a simplifying realization. We know the three particular
solutions to the following differential equations. We know the solution to the
second derivative minus 3 times the first derivative
minus 4y. Well, this is this the
homogeneous, right? And we know that the solution to
the homogeneous equation-- we did this a bunch of times--
is C1e to the 4x plus C2e to the minus x. We know the solution to-- and
I'll switch colors, just for a variety-- y prime prime minus 3y
prime minus 4y is equal to just this alone: 3e to the 2x. And we saw that that particular
solution there, y particular, was minus
1/2 e to the 2x. And we did this using
undetermined coefficients. We did that a couple
of videos ago. And then let me just write this
out a couple of times. We know the solution to
this one, as well. This was another particular
solution we found. I think it was two videos ago. And we found that the particular
solution in this case-- and this was a fairly
hairy problem-- was minus 5/17 x plus 3/17. Sorry. The particular solution was
minus 5/17 sine of x plus 3/17 cosine of x. And then finally this last
polynomial, we could call it. We know the solution when that
was just the right-hand side. That was this equation. And there we figured out-- and
this was in the last video. We figured out that the
particular solution in this case was minus x squared
plus 3/2 x minus 13/8. So we know the particular
solution when 0's on the right-hand side. We know it when just 3e to the
2x is on the right-hand side. We know it just when 2 sine of
x is on the right-hand side. And we know it just when
4x squared is on the right-hand side. First of all, the particular
solution to this nonhomogeneous equation, we
could just take the sum of the three particular solutions. And that makes sense, right? Because one of the particular
solutions, like this one when you put it on the left-hand
side, it will just equal this term. This particular solution, when
you put it in the left-hand side, will equal this term. And finally, this particular
solution, when you put it on the left-hand side, will
equal the 4x squared. And then you could add the
homogeneous solution to that. You put it in this side
and you'll get 0. So it won't change the
right-hand side. And then you will have the most
general solution because you have these two constants
that you can solve for depending on your initial
condition. So the solution to this
seemingly hairy differential equation is really just the sum
of these four solutions. Let me clean up some space
because I want everything to be on the board at
the same time. So the solution is going
to be-- well, I want that to be deleted. I'll do it in baby blue. Is going to be the solution to
the homogeneous C1e to the 4x plus C2e to the minus x
minus 1/2 e the to 2x. And I'll continue this line. Minus 5/17 sine of x plus 3/17
cosine of x minus x squared plus 3/2 x minus 13/8. And it seems daunting. When you saw this, it probably
looked daunting. This solution, if I told you
this was a solution and you didn't know how to do
undetermined coefficients, you're like, oh, I would never
be able to figure out something like that. But the important realization is
that you just have to find the particular solutions for
each of these terms and then sum them up. And then add them to the general
solution for the homogeneous equation,
if this was a 0 on the right-hand side. And then you get the general
solution for this fairly intimidating-looking second
order linear nonhomogeneous differential equation with
constant coefficients. See you in the next video, where
we'll start learning another method for solving
nonhomogeneous equations.