# Solving a differential equation w/ undetermined coefficients

1. Feb 28, 2013

### Aerospace93

1. The problem statement, all variables and given/known data
y''+2y'+y=xe-x

2. Relevant equations

Yc=c1e-x+c2xe-x

relevant info on textbook: "If any term of yp is a solution of the complementary equation, multiply yp by x (or by x2 if necessary)."
>> i dont understand the part where it says "a solution of the complementary equation". Does it mean that if any term in yp is equal to any term in the complementry solution[...]? or does it explicitly mean if any term in yp is equal to the entire solution of the complementary equation[...]?

3. The attempt at a solution
The first particular solution that pops when G(x)=xe-x is yp=(Ax+b)e-x. However, since that gives a particular equation whose terms are in the complementary equation we must multiply by x.

seconds attempt: yp=x(Ax+b)e-x.

Now what i didn't understand above comes into play. There is a term in the particular solution bxe-x which is equal to another SINGLE TERM (maybe what they refer to as a solution???) in the complementary solution c2xe-x. Does this mean i must multiply by x2?

that would be: yp=x2(Ax+b)e-x... now no single term in the particular sol. is equal to a single term in the complementary solution.

disclaimer: my confusion is with the language used. when they refer to a solution of the complementary equation are they refering to any one term in the equation and calling that a solution or are they referring to the complementary eq. as a whole? i'm confused :s and my instincts tell me that they could have used "term" instead of "solution" there too... fml

2. Feb 28, 2013

### LCKurtz

The complementary solution $y_c$ refers to the general solution of the homogeneous equation, in this case your $Y_c=C_1e^{-x}+C_2xe^{-x}$. A particular solution is any solution $y_p$ of the non-homogeneous equation.

To answer your question about the choice for $y_p$, yes, you need to multiply by $x^2$.

3. Feb 28, 2013

### Aerospace93

that's wasnt my question. to put it in other words, why would you multiply it by x2...

4. Feb 28, 2013

### LCKurtz

Here's why:

Your equation can be written in the operator form$$(D^2+2D+1)y=(D+1)^2y=xe^{-x}$$Since $y = xe^{-x}$ itself is a solution of$$(D+1)^2y =(D^2+2D+1)y = 0$$that means that if we apply another$(D+1)^2$ to both sides of this equation we get$$(D+1)^4y=(D+1)^2(xe^{-x})= 0$$The $y$ you are looking for must satisfy this equation. The general solution of $(D+1)^4y=0$ is $$y = Ae^{-x} +Bxe^{-x} +Cx^2e^{-x} +Dx^3e^{-x}$$Your particular solution must be in there somewhere, and it can't use the first two terms because they are solutions of the homogeneous equation. So for your particular solution you must look in the last two terms:$$y_p=Cx^2e^{-x} +Dx^3e^{-x}$$which is the same thing as multiplying your initial $y_c$ by $x^2$. This is the method of annihilators which is the theoretical underpinning of undetermined coefficients. There are many resources on the internet about it.

5. Feb 28, 2013

### Aerospace93

Without knowing about the form of the complementary solution at first, we'd figure that if G(x)=x2ekx then the particular solution should be something like this: (Ax2+Bx+c)ekx?? And say that you then worked out the complementary solution and it was: c1er1x+c2er1x, then the last term "c" [cekx,in the particular solution], is a solution of the complementray equation because we have two terms with (constant coefficient)*(ekx). So therefore must we multiply the particular solution by x and result w/:
yp=x(Ax2+Bx+c)ekx??

I think this is the best way for me to explain my first question.

Until now i have only learnt how to solve nonhomogeneous linear equations using the method of undetermined coefficients. So what i do is i figure the complementary then the particular and add them. Now im learning how i need to multiply the particular by x in some situation however i havent quite yet figured the logic behind it yet. I would really apreciate any help here!!

6. Feb 28, 2013

### LCKurtz

My explanation in post #4 is the logic behind it. Give it some thought. Otherwise you will be stuck with ad-hoc and confusing rules about when to multiply by what power of $x$ or what form to use for $y_p$.