# Method of Undetermined Coeffecients - how to assume the form of third degree equation.

An example differential equations questions asks me to solve;

$$y''' - 2y'' -4y'+8y = 6xe^{2x}$$

I begun by solving the homogeneous equation with $m^3 - 2m^2 -4m+8 =0$ and getting the answer $$y(x) = c_1e^{2x} + c_2xe^{2x}+c_3e^{-2x}$$

The second part of the solution involves assuming a form for the solution. Because $g(x)$ is $6xe^{2x}$, I assumed the solution would be of the form $(Ax+B)e^{2x}$, however it turns out that after differentiating three times it gets extremely complicated. Is there a better way?

Also, the textbook solutions manual uses the form of $(Ax^3 + Bx^2)e^{2x}$. How did it arrive at that? (there's no accompanying explanation)

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See here for related problem. –  Mhenni Benghorbal Nov 19 '12 at 4:56
You mean differentiating three times* instead of integrating three times? –  diimension Nov 19 '12 at 9:12
Yes thanks @diimension. I fixed it –  Imray Nov 19 '12 at 15:35
@Imray: Did you see my answer at last? –  Babak S. Dec 6 '12 at 21:36
Yes thank you! I spent a lot of time grappling over it with my textbook, pencil and some paper and now your answer makes even more sense. Thanks so much! –  Imray Dec 7 '12 at 1:34

Your equation is $y'''-2y''-4y'+8y=6xe^{2x}$. Now change the $y'$ to $Dy$ form as follows. So $$y'''\to D^3y,\\ y''\to D^2y, \; \; \text{and} \;\;y'\to Dy,$$ So by new arranging respect to $D$ operator we get our equation as: $$D^3y-2D^2y-4Dy+8y=6xe^{2x}$$ or by factoring and expanding $$(D^3-2D^2-4D+8)y=(D-2)^2(D+2)y=6e^{2x}$$ which you got before. Note that considering the corresponding homogeneous equation $$(D-2)^2(D+2)y=0$$ we get $(D-2)^2=0,\;\; (D+2)=0$ which leads us to write the general solutions as $$y_c(x) = c_1e^{2x} + c_2xe^{2x}+c_3e^{-2x}$$ and you did it right above before. Now have a look at some facts:

• If $y=\text{constant}$ so $y'=0$ or $Dy=0$. Here, the operator $D$ annihilates $y$ which is just a constant.($Dc=0$)
• If $y=cx$ in which $c$ is a constant so $y''=0$ or $D^2y=0$. It means that the operator polynomial $P(D)=D^2$ annihilates $y=cx$. $(\text{or} \; P(D)=D^2(cx)=cD^2x=c(x'')=0)$. Generally, $D^{n+1}$ annihilates not only the function $y=cx^{n}$ but also all linear functions as $$y=c_0+c_1x+c_2x^2+...+c_nx^n$$ It means that $$P(D)y=D^{n+1}y=0$$.
• As the same the differential operator $(D-\alpha)^n$ annihilates each of the following functions and every linear combinations of them: $$e^{\alpha x},xe^{\alpha x},x^2e^{\alpha x},...,x^{n-1}e^{\alpha x}$$ Now look at the RHS of your original equation. I mean $=6xe^{2x}$. Can we guess of which proper differential polynomial annihilates it? As above it would be $(D-2)^2$. It means that $(D-2)^2 \left(6xe^{2x}\right)=0$. Don't respect to numeric coefficients like $6$ here at all.

Now we consider of what we have achieved at last: $$(D-2)^2(D+2)y=6e^{2x}$$ Put the operator $(D-2)^2$ before both sides of the above converted equation: $$(D-2)^2\left((D-2)^2(D+2)y\right)=(D-2)^26e^{2x}=0$$ or $$(D-2)^4(D+2)y=0$$ In fact, we have found a proper differential operator $P(D)=(D-2)^4(D+2)$ which if it effects to $y$, $y$ will be lost.

Now, for a while, forget our equation and look at $(D-2)^4(D+2)y=0$ and think someone gave this to us asking to guess which function $y$ may satisfy the equality above? We reply:

• Since we have $(D-2)$ so we have some forms as $e^{2x}$ in $y$.
• Since $(D-2)$ has a power $4$ so we have the forms $Ae^{2x},\; Bxe^{2x}, \;Cx^2e^{2x}, \text{and} \; Ex^3e^{2x}$ in $y$. Note that you multiply $e^{2x}$, in the previous line, by $A,\; Bx,\; Cx^2,\; Ex^3$. (Exactly until the power of $x$ gets $4-1=3$).

• And, since we have $(D+2)$, then $y$ has the term $Fe^{-2x}$.

So we are done. Our probable function which satisfy the original equation is $$y=Ae^{2x}+ Bxe^{2x}+Cx^2e^{2x}+Ex^3e^{2x}+Fe^{-2x}$$. Now put the terms which generate $y_c(x)$ aside and take the rest for what we have looked for. It is $$y_p=Cx^2e^{2x}+Ex^3e^{2x}$$ where $C,E$ are unknown constant.

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What does $P(D)y=(D-2)^4(D+2)y=0$ have to do with $y_p=Ax^2e^{2x}+Bx^3e^{2x}$? –  Imray Nov 19 '12 at 2:17
Thanks Babak, sure I can wait until tomorrow. I really appreciate it! –  Imray Nov 19 '12 at 18:03

The correct form for the inhomogeneous solution is $(Ax^3 + Bx^2) e^{2x}$, so your solution manual is correct. The general strategy for undetermined coefficients is as follows:

1. Write down the homogeneous solution $y_h(x)$ by finding the roots $m$ of the auxiliary equation (you did this right).

2. Look at the inhomogeneous term, which is of the form $P(x) e^{rx}$, where $P(x)$ is a polynomial of degree $d$. In this case, it was $P(x) = 6x$ and $r = 2$.

3. Write down the "guessed" form of the inhomogeneous solution, which will be of the form $$Q(x) x^k e^{rx}$$ where $Q(x)$ is an undetermined polynomial of degree $d$ (so $Q(x) = Ax + B$ in your case), and $k$ is the multiplicity of $r$ as a root of the auxiliary equation. Here, since $r = 2$, it is a double root of the auxiliary equation, so we get an additional $x^2$ term.

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Ok, I understand where the $P(x)$ comes from - there's a polynomial in the $g(x)$. But how did you think of putting the $Q(x)$ in? –  Imray Nov 19 '12 at 2:16
The heuristic for undetermined coefficients is that the form for the inhomogeneous solution must be similar to that of the inhomogeneous term in the equation. So if the inhomogeneous term is polynomial times exponential, we "guess" that our solution is also polynomial (of equal degree) times exponential. Then, in order to account for multiple roots, we have to multiply by additional powers of $x$. –  Christopher A. Wong Nov 20 '12 at 22:30
But $6x$ is a polynomial of highest degree $1$ so why do we go up to $x^3$? –  Imray Nov 21 '12 at 1:12