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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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    @JoelReyesNoche Earlier question: [link](http://math.stackexchange.com/questions/112583/complementary-solution-homogenous-solution/112635#112635)2012-02-24

4 Answers 4

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$r=1$ 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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    Thanks +1, I have to practise this as I can know understand the term `"multiplicity"`.2012-02-24
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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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    @hhh: Thanks for spotting the typo.2012-02-27
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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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    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.2012-02-24