# $y'''-y=x^{2}$ has solution — “multiplicity”?

The page 667 of the book (sorry not in English) claims $y'''-y=x^{2}$ to have the solution

$$y(x)=C_{1}e^{x}+e^{-x/2}\left(C_{2} \cos \left( \frac{\sqrt{3}x}{2} \right)+C_{3} \sin\left(\frac{\sqrt{3} x}{2}\right)\right) -x^{2}.$$

The book mentions that with the $m$ -multiple real root solution is $x^{k}e^{rx}$ and with $m$ -multiple conjugate pair $\alpha\pm i\beta$ the solution is $x^{k}e^{\alpha x}\cos(\beta x), x^{k} e^{\alpha x} \sin(\beta x)$ where $k=0...m-1$.

Let's check how to use it in this example. We have 3th order DY so $m=3$. But what kind of roots does this have $r^{3}-1=0$?

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$r^3-1=(r-1)(r^2+r+1)$; which has the roots $1$ and ${-1\pm i\sqrt 3\over 2}$. – David Mitra Feb 24 '12 at 13:34
There are no multiple roots (see David Mitra's comment), so you won't be needing those $x^k$ factors in the general solution of the homogeneous DE. – Jyrki Lahtonen Feb 24 '12 at 13:38
There are three roots: $r=1$ with multiplicity $m=1$; and the conjugate pair $-.5\pm.5\sqrt3 i$ with multiplicity $m=1$. This gives the solution to the homogeneous equation: $C_1 e^x+e^{-x/2}(C_2 \cos (\sqrt3x/2)+C_3\sin(\sqrt3x/2))$. The "m"'s being refered to in the notes are the multiplicities of the roots; not the degree of the ce. For a root with multiplicity $1$, there is no "$x^k$"-factor. – David Mitra Feb 24 '12 at 14:04
@DavidMitra: ...what does the term "multiplicity" mean? I start to see what it means, ti must be something related to multiple roots with polynomials here (not the degree of the DYs)? – hhh Feb 24 '12 at 14:05
Yes, the link you mention explains it. Example: $(x-2)^3 (x-1)^2(x+1)$. $2$ is a root with multiplicity 3, 1 is a root with multiplicity 2, $-1$ is a root with multiplicity 1. – David Mitra Feb 24 '12 at 14:08

## 4 Answers

$r=1$ is a solution of $r^3-1$. So $$r^3-1=(r-1)q(x).$$ for some quadratic expression $q$. Doing the division $q(x)={r^3-1\over r-1}$ yields $$r^3-1=(r-1)(r^2+r+1).$$ (Or, use the difference of cubes formula on $r^3-1^3$).

The quadratic formula gives the roots of $r^2+r+1$: $r={-1\over 2}\pm{\sqrt 3\,i\over 2}$.

So, the three roots are: $1$, ${-1\over 2}+{\sqrt 3\,i\over 2}$, and ${-1\over 2}-{\sqrt 3\,i\over 2}$.

You do not have $m=3$ here for the formula you gave. In that formula, $m$ depends on which root you are looking at and refers to the power of the corresponding factor occurring in the characteristic equation.

Here, the characteristic equation is $$r^3-1 =(r-1)^\color{red}1 \textstyle\bigl( r- ({-1\over 2}+{\sqrt 3\,i\over 2})\bigr)^\color{green}1\bigl( r- ({-1\over 2}-{\sqrt 3\,i\over 2})\bigr)^\color{green}1.$$

For the root $r=1$, we have $m=\color {red}1$. (Which gives the term $C_1 e^x$ in the solution. Note, with $m=1$, the solution formula gives only one term: $x^0e^{rx}=e^{rx}$.)

For the complex conjugate pair root $r={-1\over 2}\pm{\sqrt 3\,i\over 2}$, we have $m=\color {green}1$. (Which gives the term $e^{-x/2} \bigl(C_2 \cos(\sqrt3x/2)+C_3\sin(\sqrt3x/2)\bigr)$ in the solution.)

As another example, suppose one had a homogeneous linear differential equation with constant coefficients that had characteristic equation: $$(r-2)^2(r+3)^3(r+1).$$

Then the roots are

$\ \ \ \ r=2$ with $m=2$

$\ \ \ \ r=-3$ with $m=3$

$\ \ \ \ r=-1$ with $m=1$

The general solution to the equation would be $$(C_1e^{2x}+C_2 xe^{2x})+( C_3e^{-3x}+C_4 xe^{-3x}+ C_5x^2e^{-3x})+C_6 e^{-x}.$$

Your formula is imprecisely stated. Here is a complete version:

First a definition:

The multiplicity of the root $r$ of the polynomial $q(x)$ is the largest integer $m$ such that $(x-r)^m$ is a factor of $q(x)$.

Now the "recipe":

For the homogeneous equation with constant coefficients: $$\tag{1}\def\sss{} a_{\sss n} y^{\sss( n )} +a_{\sss n - 1} y^{\sss( n - 1 )} +\cdots+a_{\sss1} y' +a_{\sss0} y = 0 , \quad a_n\ne0,$$
the associated characteristic polynomial (c.p., henceforth) is $$\tag{2}\def\sss{} a_{\sss n}x^n+a_{\sss n\!-\!1}x^{n-1} +\cdots+a_{\sss1}x +a_{\sss0} .$$

To find the general solution of equation $(1)$: You want to first find a set of $n$, independent, solutions to equation $(1)$. Then you form the general solution by writing it as a general linear combination of the $n$, independent, solutions found.

Towards this end, you may:

1. Find the roots and their corresponding multiplicities of the c.p. $(2)$. Complex roots will occur as complex conjugate pairs. We will then speak of the "complex conjugate root $a\pm bi$ with multiplicity $k$", whose meaning is, hopefully, evident.
2. Note: If $c$ is a real root of $(2)$ with multiplicity $k$, then $k$-independent solutions of $(1)$ are $$e^{ct},\ xe^{ct},\ x^2 e^{ct},\ \ldots,\ x^{k-1}e^{ct}.$$Note that for $k=1$, there is only one term: $e^{ct}$.
3. Note: If $a\pm bi$ is a complex conjugate pair root of $(2)$ with multiplicity $k$, then $2k$-independent solutions of $(1)$ are $$e^{at}\sin (bt),\ x e^{at}\sin (bt),\ \ldots,\ x^{k-1} e^{at}\sin (bt)$$ $$e^{at}\cos (bt),\ xe^{at}\cos (bt)\ ,\ \ldots,\ x^{k-1}e^{at}\cos (bt).$$Note that for $k=1$, there are only two terms: $e^{at}\sin(bt)$ and $e^{at}\cos(bt)$.
4. Write down all solutions given by steps 2. and 3.: For each real root of the c.p., list the solutions given by step 2; and, for each complex conjugate pair root, list the solutions given by step 3. This will generate a set of $n$ independent solutions to equation $(1)$. The general solution to $(1)$ is then

$$y_c=c_{\sss 1}y_{\sss1}+c_{\sss2}y_{\sss2}+\cdots+c_{\sss n}y_{\sss n}$$ where $y_1$, $y_2$, $\ldots\,$, $y_n$ are the $n$-solutions found above.

Once you've developed some facility with this, you should be able to just write down the solution by looking at the fully factored c.p.

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It may be worth reiterating that each of those three roots have multiplicity 1 (m=1, so k=0 only). – Andrew Parker Feb 24 '12 at 14:22
Thanks +1, I have to practise this as I can know understand the term "multiplicity". – hhh Feb 24 '12 at 14:34

Your current question has a characteristic polynomial of degree 3. This means that there are 3 roots, when you count multiplicities. In general, let $d$ be the degree of a polynomial $f(x)$, let $r_1, r_2, \ldots, r_n$ be the distinct roots of $f(x)$ with respective multiplicities $m_1, m_2, \ldots, m_n$. Then $d = \Sigma m_i$.

Since we find three roots (1, $(1+i\sqrt{3})/2$, and $(1-i\sqrt{3})/2$), and their multiplicities must add up to 3 (the degree of our characteristic polynomial), the multiplicity of each root must be 1. Thus we do not need more than k=0 for each root.

If you're looking for an example of a differential equation that has $m \gt 1$, you should look at $y'' - 2y' + 1 = 0$. The characteristic polynomial of this equation has the same root repeated twice, $(r-1)(r-1)$, meaning that the root $r=1$ has multiplicity 2. Thus the homogeneous solution is $y=C_1e^x+C_2xe^x$

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The roots of the characteristic equation $r^3-1=$ are $r=1$ and $r=(-1\pm i\sqrt3)/2$, so the general solution of the homogeneous DE $y'''-y=0$ is $$y=C_1e^x+e^{-x/2}\left(C_2\cos\left(\frac{\sqrt{3}}{2} x\right)+C_3\sin\left(\frac{\sqrt{3}}{2} x\right)\right).$$ The usual ansatz for finding a single solution of the non-homogeneous DE works, so I don't see what problems remain?

You seem to be confusing the degree of the equation with the multiplicity of a root?

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+1 for the Mittag-Leffler functions. – draks ... Feb 24 '12 at 15:45
sorry about the typo (sign error) in the real parts of the complex conjugate roots. The solution (of the homogeneous DE) was ok. – Jyrki Lahtonen Feb 24 '12 at 16:47
@hhh: Thanks for spotting the typo. – Jyrki Lahtonen Feb 27 '12 at 6:38

EDIT: What follows is relevant to the question as originally posted. The question was edited after I posted this answer, and this answer is no longer relevant to the revised version of the question.

It's the $m=2$ case in your notation, so you need $x^ke^{\alpha x}\cos(\beta x)$ and $x^ke^{\alpha x}\sin(\beta x)$ for $k=0..1$.

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Do you infer the below for the solution? $$y(x)=A+1 (B cos(x)+C sin(x)) +1 (D cos(-x)+E sin(-x))+ \left(F e^{\alpha x}cos(\beta x)+G e^{\gamma x} sin(zx)\right) + \left(Hx e^{tx} cos(vx) + Ix e^{ux} sin(wx)\right)$$ – hhh Feb 24 '12 at 12:19
I can't read your comment - the TeX isn't TeXing - but I'm using the same notation you use in the sentence that begins, "The book mentions," and what I wrote is what I meant. – Gerry Myerson Feb 24 '12 at 12:22
Now I can read your comment, and I have no idea what your $\gamma$ and $z$ and $t$ and $v$ and $u$ and $w$ are supposed to be. I wrote something fairly simple, and you are doing something way complicated. Just try to do what I wrote. – Gerry Myerson Feb 24 '12 at 12:25