Dif Eq - Undetermined Coefficients

1. Oct 27, 2008

Nubcakes

This isn't really a homework question, but it's a question from my book so I thought this would probably be the appropriate forum to ask how to do these types of problems. As the title says this is a Differential Equation of Undetermined coefficients. I just need to solve it's general solution so to speak;

NOTE: Please forgive me for my god awful on-screen writing. :uhh:

I've asked a few of my professors about this and they ALL seem to have different approaches and none of them like to show all of their steps. So, I attempted to make sense of it using the least confusing method one of them showed me, but I am unsure if I did the problem correctly;

2. Oct 27, 2008

rock.freak667

Your finding of the complimentary solution is correct.
Finding the particular integral was incorrect y_p=Ax+B, but when you put that into the equation you'd get

-10(A)+25(Ax+B)=30x+3

Meaning that if you equation coefficients

25Ax=30x =>A=30/25

and 25B-10A=3

3. Oct 27, 2008

Staff: Mentor

I'm a bit rusty on this, but I think I can explain what's going on.

IF the DE were y'' - 10y' + 25y = 0, you know that the solution would be y = c1e^(5x) + c2xe^(5x). This is the solution of what's called the homogeneous equation, so let's call this yh.

Now the differential equation you're actually trying to solve has a polynomial on the right side, so the tactic is to "guess" a solution of the form yp = Ax + B, where the p subscript denotes particular.

The work you showed is a little disorganized, and you came out with the wrong values for A and B.

Since yp is supposed to be a solution of the DE y'' - 10y' + 25y= 30x + 3, let's substitute yp in this DE and determine these undetermined coefficients (A and B).

yp = Ax + B
yp' = A
yp'' = 0

So yp'' - 10 yp' + 25 yp = 30x + 3
==> 0 -10A + 25(Ax + B) = 30x + 3
==> 25Ax + 25B -10A = 30x + 3

Since the equation above has to be true for all x, it must be that the coefficient of x on the left equals the coefficient of x on the right, and the constant terms on both sides must be equal as well.

We see that 25A = 30 and 25B - 10A = 3
==> A = 6/5 and B = 3/5.

This means that the particular solution, yp, is 6/5 *x + 3/5.

The general solution consists of all of the solutions to the homogeneous equation plus the particular solution to the equation with the 30x + 3 on the right.

The general solution, therefore, is yg = yh + yp, or
c1e^(5x) + c2xe^(5x) + 6/5 x + 3/5

If you think about it with the two separate parts of the general solution, the solution to the homogeneous equation (yh) is such that if you take its 2nd derivative, subtract off 10 times its first derivative, and add 25 times it, you get 0. If you apply the same operations to the particular solution, yp, you end up with 30x + 3.

The basic idea with a nonhomogeneous equation like this one is that the general solution will be all of the solutions to the homogeneous equation (y'' + ... = 0) plus a particular solution to the nonhomogeneous equation.

Does that help?

4. Oct 27, 2008

Nubcakes

Thanks alot Rock.freak667 and Mark44!

Both of you helped quite abit! Before I thought that the constants out infront on the Y side were simply multiplied to the Ax+b y_p equations and you simply solved for the unknown constants. From there I assumed you just slap them into the Y = equation with the homogen part I got from the Y side.

If I understood what you said, the core part where I went wrong is in the red box below,

Which should look something like this;

Sorry for my horrible writing and disorganization!

Anyway, if what I just showed is correct then it looks like I understood what you both said.

EDIT: I should have added; I think A has X attached to it in the equation because A is directly multiplied to X^1 in the first term it is seen in where B is just multiplied by a constant