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### Course: Differential equations>Unit 2

Lesson 3: Method of undetermined coefficients

# Undetermined coefficients 1

Using the method of undetermined coefficients to solve nonhomogeneous linear differential equations. Created by Sal Khan.

## Want to join the conversation?

• Where might I find a video on the Annihilator Approach? I understand that the complete solution is the sum of the homogeneous solution and the particular solution, but I would like to better understand how to use the annihilator method to find solutions.
• I really don't quite get why we need to go about finding a general solution to a nonhomogeneous equation, when I already have an answer that works in the particular solution. I think this needs to be explained or at least talked about, and I'm not just referring to this site but rather ALL texts I have read on DE (and that is a LOT). Why would I want extra parameters to map out all possible solutions when I am solving a specific DE? Can someone provide an example of how and when this would be helpful, i.e. when a particular solution is not enough?
• The particular solution is a solution but it's not the only solution.
So in addition y = -1/2 e^2x being a solution you also could have a solution like:
y = 3 e^4x + 2 e^-x - 1/2 e^2x
(and you can verify that's a solution)

So in order to express all solutions you need to include the homogeneous solution with the arbitrary constants. It's similar to integral calculus where you need to include the "+ C" on the end of your solution to express all possible solutions. The particular solution is not like an initial value problem where you are finding the one and only answer.
• How do we know that there are not any other general (edited) solutions to this differential equation?
• Differential equations in general have a whole class of solutions, each making the equality true. In the inhomogeneous linear case every solution may be expressed as a sum of an arbitrary solution to the inhomogeneous equation plus a solution to the associated homogeneous equation.

You may find a proof that such sums cover all the possible solutions in the 12th lecture of the 18.03 course: http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/undetermined-coefficients/. The part of the video from 17' to 28' also includes a bit more general demonstration of what Sal shows in his video -- that such sums actually are solutions -- as well as an explanation of the "particular solution" language.

When looking for a particular solution (solution to the original, inhomogeneous equation) it's enough to find one, no matter which one. In the example we could have guessed e^(-x) - 1/2 e^(2x), checked that this is in fact a solution, and used that to write all solutions (by adding the same parametrized homogeneous solution). Nevertheless, finding even one solution doesn't always turn up to be as easy as it may sound.
• Anyone know where I can get a good tutorial explaining what "homogeneous" is?
• The formal definition is: f(x) is homogeneous if f(x.t) = t^k . f(x), where k is a real number. It means that a function is homogeneous if, by changing its variable, it results in a new function proportional to the original. By this definition, f(x) = 0 and f(x) = constant are homogeneous, though not the only ones.
• What if in the original equation, instead of =g(x) I have =5 (or any other constant)?
• This is, in fact, still of the form g(x) where g(x) = 5 + 0*x.
• In case a member of the family of derivatives of the function is featured in the solution to the homogenous equation, we multiply by x till we get rid of that problem. Could someone explain why we do so?
• What if g(x) is a constant B. I know you can solve it using the perticular solution and homogenous solution, but when you solve it without doing that you get:
e^ax(a^2 + ba + (c-B)) = 0. Is it possible to solve it this way, and if it isn't why not?
• If g(x) is a constant then your particular solution should be a constant. y'' and y' will be 0 so you're just solving the equation c*y = B
(1 vote)
• what is Yp for D^2(y)+ 2Dy = 8x + e^(-2x);
(1 vote)
• Fundamental set: {1, e^-2x}
We guess Yp1: Ax + B, but this contains an element already in the fundamental set, so we multiply by x, and then the correct form of Yp1 is Ax^2 + Bx.
We guess Yp2: Ce^-2x. but this is already contained in the fundamental set, so we multiply by x, so the correct guess for Yp2 is Cx*e^-x.
So the form of Yp: Yp=Ax^2 + Bx + Cx*e^-2x. after taking derivatives, and equating coefficients, you will find A = 2, B = -2, C = -1/2
• @ ish, what happens when the left side all adds up to 0?
ex: y''(x)-6y'(x)+9y(x)=5e^3x is the equation and my guess would be Ae^3x
i'd end up with coefficients of 9-18+9=5e^3x
• In that case, you've proved your guess wrong by contradiction. That means you must change your guess accordingly. For example, I would change y(x) = Ae^3x to y(x) = Ae^Bx instead.

This gives y(x) = Ae^Bx, y'(x) = ABe^Bx and y''(x) = AB^2e^3x
y''(x) - 6y'(x) + 9y(x) = AB^2e^3x - 6ABe^3x + 9Ae^3x = 5e^3x
Therefore, AB^2 - 6AB + 9A = 5.

Solving for A, we have A = 5/(B - 3)^2; B =/= 3

That's it. Hope that helps!
(1 vote)
• can we have multiple answer for the same solution base on different guess ?