Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $r^{3}-1=0$?

share|cite|improve this question
$r^3-1=(r-1)(r^2+r+1)$; 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
up vote 5 down vote accepted

$r=1$ 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.

share|cite|improve this answer
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 $y=C_1e^x+C_2xe^x$

share|cite|improve this answer

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?

share|cite|improve this answer
+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 $k=0..1$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.