Main content
Course: Differential equations > Unit 2
Lesson 3: Method of undetermined coefficientsUndetermined coefficients 4
Putting it all together! Created by Sal Khan.
Want to join the conversation?
Hello! I'm an engineering student and had an exam with an undetermined coefficients exercise I couldn't solve using this method. Can you help me noticing where I went wrong?
Problem was: y'' + 4y = sin(2x)
I went ahead and predicted the solution in the form of Asin(2x) + Bcos(2x). I figured the 1st and 2nd derivative, and replaced them in the equation:
(-4Asin(2x) - 4Bcos(2x)) + 4(Asin(2x)+Bsin(2x)) = 1sin(2x) + 0cos(2x)
Since all the A's and B's even out, I can't get past this. Thanks!(12 votes)- Always start by solving the homogeneous equation of the differential equation;y'' + 4y = 0 which has solutions of y_h = c_1cos(2x) + c_2 sin(2x)
Then start by assuming the particular solution as you mentioned:
y_p = Asin(2x) + Bcos(2x),
however this will not work because both predicted solutions are homogeneous solution. Similar to when you have repeated roots you need to multiply each function by x, so the predicted solution should be
y_p = Axsin(2x) + Bxcos(2x)(26 votes)
How about the topic of variation of parameters?(23 votes)
what if we had (e^2x)*(sin2x)*(4x^2) on RHS?(10 votes)
I guess you’d have to try y_p(x) = A * x^2 * sin(2x)* e^(2x) + B * x^2 * cos(2x)* e^(2x) + C * x * sin(2x)* e^(2x) + D * x * cos(2x)* e^(2x) + E * sin(2x)* e^(2x) + F * cos(2x)* e^(2x), though I admit that I’m too lazy to check. Anyone?
Or just use a CAS: http://www.wolframalpha.com/input/?i=y%27%27+-3y%27+-4y+%3D+e%5E%282x%29*sin%282x%29*4x%5E2(6 votes)
- Hi I am a first year Applied Mathematics student. This is really helpful so thanks so much for that. I was really wondering what you would do if you had xsinx on the right hand side as opposed to simply some constant multiplied by the sinx? Thanks so much. This is coming from South Africa(4 votes)
- In that case the easiest way to go about it is:
Realize that x is a polynomial like X^2
So the solution for a differential equation for it will be X+C
and for sin x you know its= Asinx+BcosX so multiply the two.
you get a function that might work . Am not sure.
On second thought
The best way to generalize it is that When you have a non-homogeneous equation the particular solution is usually A constant times the function + another Constant times the function's derivative+another other constant times the next derivative until you reach a constant-0- or until you stop getting new euaton forms like in trig. functions then combine like terms and turn any constant you might get to letters. This is the general form of the particular equation And if you look back that rule works.
so for Y=xsinx
Y'=sinx-xcosx
y''=cosx-cosx+xsinx
you can keep deriving for fun, combine the like terms and you will see that the particular solution is of the form : Axsinx+Bxcosx+Dsinx+Ecosx
Hopefully that works. i should go try it out and get back....
LONG WINDED! SORRY!(10 votes)
Where can I find the "next video" with another method for solving nonhomogenous d.e. mentioned in5:47?(6 votes)- That would be the set of videos on Laplace Transforms. Check them out under the Differential Equations topic page.(3 votes)
how we can integrate e^x/1+e^2x(1 vote)- You need to use trigonometric substitution, to know what trigonometric substitution to use you need some practice, since this integral has the form
a + x^2, we need to use the following substitution:e^x = tan(θ)
And we need to get the differential of this substitution, so we derivate both sides:e^x dx = sec^2(θ) dθ
And now we substitute this into the integral:⌠ e^x ⌠ sec^2(θ) ⌠ sec^2(θ)
| ―――――――――― dx = | ―――――――――――― dθ = | ―――――――― dθ = ∫dθ = θ + C
⌡ 1 + e^(2x) ⌡ 1 + tan^2(θ) ⌡ sec^2(θ)
After using the trigonometric identity1 + tan^2(θ) = sec^2(θ)the integral simplifies a lot, and we get our first resultθ + C.
Now we need to go back to our original substitution and solve forθ.e^x = tan(θ)
arctan(e^x) = arctan(tan(θ))
arctan(e^x) = θ
And with this we get the final solution:⌠ e^x
| ―――――――――― dx = arctan(e^x) + C
⌡ 1 + e^(2x)(5 votes)
- What if we were to go simple, such as: y''-10y'+25y=30x+3?? I end up with three variables and i do not know of a way to solve it.(1 vote)
- General solution is:
Yg = C1*e^(5x)
Consider that the particular is a polynomial of the form:
Yp = Ax + B
Yp' = A
Yp" = 0
0 - 10A + 25(Ax+B) = 30x + 3
25Ax -10A+25B = 30x + 3
Therefore you get two equations with two variables:
| 25A = 30
| -10A+25B = 3
solved is...
| A = 30/25 = 6/5
| B = 3/5
So the particular solution is...
Yp = (6/5)x + 3/5(5 votes)
- So after watching the whole series on undetermined coefficients, I'm still stumped by (and this may be silly) the special case where g(x) is equal to a constant (not a constant function, just a constant). What form does the particular solution take when the equation looks like y'' - 3y' -4y = 6?
I have a feeling the answer is very simple, but I can't find a concrete answer. Thank you in advance!(2 votes)
Your particular solution would just take the form A. So then y' = 0 and y''=0. Then, plugging in y'', y', and y into y''-3y'-4y, we get -4A = 6. Thus A= -3/2 or -1.5. Your general solution would come be y(t) = c1 * e^(-t) + c2* e^(4t). Putting both solutions together, we get y(t) = c1 * e^(-t) + c2* e^(4t) - 1.5.
Hope this helps.(2 votes)
- how we can solve D(D-1)y+Dy+y=Sin(logx^2)(1 vote)
- Hi! What if g(x)=tan(k*x)? Seems not so easy to predict 'cause of the form of the derivative of tan(kx)(2 votes)
5:47Video 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.