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Let $ \sum\limits_{k = 0}^n {y^{\left( k \right)} a_k } = 0 $ an homogeneous ODE, where $a_k$ are constants. How can I solve the equation when the roots are repeated? One way, that I saw in wikipedia, is using the fact that if $ e^{cx} $ is a solution, then $ (x^r)(e^{cx}) $ is also too, How can I prove this? It´s difficult to me, to evaluate the sum, because i want to show that $ \sum\limits_{k = 0}^n {\left( {x^r e^{cx} } \right)^{\left( k \right)} a_k } = 0 $ but I need to evaluate $ {\left( {x^r e^{cx} } \right)^{\left( k \right)} } $ and I don´t know how to do it Dx

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The idea is to perturb coefficients of your equation a little so as to make all the roots distinct, and then consider the limit as the amount of perturbation goes to zero. – Sasha Nov 6 '11 at 20:56

I don't know if you have the background for this, but I do not think there is much choice if you allow the degree $n \geq 3.$ It is not so bad for $n=2.$

Anyway, you introduce a bunch of variables $y_0 = y, y_1 = y', y_2 = y'',$ and so on. As your coefficients are constant, we may divide through by whatever $a_n$ might be to arrive at a revised constant coefficient equation with $a_n = 1.$

So the new system of equations is a linear system beginning with $$y_0' = y_1, \; y_1' = y_2, \ldots, \; y_{n-2}' = y_{n-1},$$ but finally $$ y_{n-1}' = - b_0 y_0 - b_1 y_1 - \ldots - b_{n-1} y_{n-1},$$ where I have taken $b_j = a_j / a_n.$

We write a column vector $Y$ with entries $y_0, y_1, \ldots, y_{n-1}.$ that system now becomes $$ Y' = B Y $$ with initial conditions written as $$ Y(0) = Y_0$$ and the solution to the system is $$ Y = e^{B x} Y_0$$

Note that $B$ has exactly the form of a companion matrix, see the square matrix at

The appearance of $x, x^2,\ldots$ comes from the Jordan normal form of $B,$ precisely when there are repeated eigenvalues (characteristic values) of $B,$ and when there are off-diagonal entries in the relevant Jordan block. If there are repeated roots but the Jordan normal form is diagonal anyway, then no polynomial terms appear. I do not know anything special about the Jordan normal form of a companion matrix, perhaps something precise can be said that need not hold for other types of coefficient matrix $B.$

This is half a semester of work if you have already had linear algebra. Anyway, see

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An alternative to looking at the corresponding first order linear system is to consider factoring linear differential operators.

Given a homogeneous linear differential equation with constant coefficients, $a_ny^{(n)}+\cdots+a_1y'+a_0y=0$, we can define a linear differential operator $L=a_nD^n+\dots+a_1D+a_0$ where $D[y]=y'$, $D^2[y]=D[D[y]]=y''$, etc. and each constant stands for left multiplication by that constant: $a_k[y]=a_ky$. So our differential equation becomes the operator equation: $L[y]=0$.

Example: $3y''+2y'+5y=0$ $\Longleftrightarrow$ $L=3D^2+2D+5$ and $L[y]=0$

Next, $D$ is a linear operator, so it commutes with constants: $D[ky]=kD[y]$. In addition, powers of $D$ commute with each other. This means we can factor each operator (like a polynomial).

Example: $y''-y=0$ gives us $L=D^2-1=(D-1)(D+1)=(D+1)(D-1)$ these are all equal because everything commutes.

Now to solve our equation we can focus on one factor at a time.

Example: Suppose $L=(D-1)(D-2)$. Then $L[e^t]=(D-1)(D-2)[e^t]=(D-2)(D-1)[e^t]=(D-2)[0]=0$ since $(D-1)[e^t]=e^t-e^t=0$ and $(D-2)[0]=0-2(0)=0$. Next, $L[e^{2t}]=(D-1)(D-2)[e^{2t}]=(D-1)[2e^{2t}-2e^{2t}]=(D-1)[0]=0$. This shows that both $y=e^t$ and $y=e^{2t}$ are solutions and hence $y=c_1e^t+c_2e^{2t}$ is the general solution (accepting some theory which tell us that 2 solutions are linearly independent and the space of solutions is 2-dimensional).

Let's focus on a single factor: $L=(D-r)^n$.

Consider $y=t^ke^{rt}$. $(D-r)[y]=(kt^{k-1}e^{rt}+t^kre^{rt})-r(t^ke^{rt})=kt^{k-1}e^{rt}$. Next, $(D-r)^2[y]=(D-r)(D-r)[y]=(D-r)[kt^{k-1}e^{rt}]=k(k-1)t^{k-2}e^{rt}$. And in general $(D-r)^{k+1}[y]=k(k-1)\cdots 2\cdot 1\cdot 0 \cdot e^{rt}=0$.

This means that $(D-r)^n[t^ke^{rt}]=0$ when $n>k$.

Thus if the characteristic polynomial has a factor $(x-r)^n$ so that the corresponding linear differential operator $L$ has a factor $(D-r)^n$, we must have solutions $e^{rt}$, $te^{rt}$, $\dots$, $t^{n-1}e^{rt}$. These can be shown to be linearly independent by using a Wronskian determinant (this is fairly straight forward but difficult for a beginner). Solutions corresponding to distinct factors are independent because they are in distinct spaces of generalized eigenvectors.

Example: Suppose $y^{(5)}-4y^{(4)}+y'''+10y''-4y'-8y=0$. Then $L=D^5-4D^4+D^3+10D^2-4D-8=(D-2)^3(D+1)^2$. Solving $(D-2)^3[y]=0$, we get $y=c_1e^{2t}+c_2te^{2t}+c_3t^2e^{2t}$. Solving $(D+1)^2[y]=0$, we get $y=c_4e^{-t}+c_5te^{-t}$. The general solution is $y=c_1e^{2t}+c_2te^{2t}+c_3t^2e^{2t}+c_4e^{-t}+c_5te^{-t}$.

Similar methods allow one to deal with (repeated) irreducible quadratic factors.

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I''l use two theorems to prove your point. One is:

THEOREM 1: If $y_1$, $y_2$, $y_3$,$\cdots$, $y_k$ are $k$ solutions of the

$$\phi(D)y = 0$$

where $\phi(D)$ is a polynomial in $D =\displaystyle \frac{d}{dx}$

then $$y = \sum_{i=i}^k c_i y_i$$

with each $c_i$ constant is a solution.

PROOF Given that $\phi(D)$ is a linear operator, if one has $$\eqalign{ & c_1\phi \left( D \right){y_1} &=& 0 \cr & c_2\phi \left( D \right){y_2} &=& 0 \cr & \cdots &=& \cdots \cr & c_k\phi \left( D \right){y_k} &=& 0 \cr} $$

then summing produces

$${c_1}\phi \left( D \right){y_1} + {c_2}\phi \left( D \right){y_2} + \cdots + {c_k}\phi \left( D \right){y_k} = 0$$

$$\phi \left( D \right)\left( {{c_1}{y_1} + {c_2}{y_2} + \cdots + {c_k}{y_k}} \right) = 0$$

So $$y = {{c_1}{y_1} + {c_2}{y_2} + \cdots + {c_k}{y_k}}$$ is a solution.


If $$(D-a)^n y =0$$ then $$y=\left(c_0+c_1 x+\cdots+c_n x^{n-1}\right)e^{ax}$$ is a solution.


It is true that $$(D-a)(e^{ax} u) = e^{ax} Du$$ since

$$ae^{ax}u+e^{ax}u'-ae^{ax}u = e^{ax} u'$$

Let's show by induction that $$(D-a)^n(e^{ax}u) = e^{ax} D^nu$$

It is true for $n=1$. But $n=k$ implies $n=k+1$, since

$$(D-a)^{n+1}(e^{ax}u) =(D-a) (e^{ax} D^nu)$$

$$\eqalign{ & {\left( {D - a} \right)^{n + 1}}\left( {{e^{ax}}u} \right) &=& D\left( {{e^{ax}}{D^n}u} \right) - a{e^{ax}}{D^n}u \cr & &=& a{e^{ax}}{D^n}u + {e^{ax}}{D^{n + 1}}u - a{e^{ax}}{D^n}u \cr & &=& {e^{ax}}{D^{n + 1}}u \cr} $$

Therefore it is true for all $n$.

Putting $y = e^{ax} u$ in our original DE produces.

$$\eqalign{ & {\left( {D - a} \right)^n}y &=& 0 \cr & {\left( {D - a} \right)^n}\left( {{e^{ax}}u} \right) &=& 0 \cr & {e^{ax}}{D^n}u &=& 0 \cr & {D^n}u &=& 0 \cr} $$

If $$u = \sum\limits_{k = 1}^n {{c_k}{x^{k - 1}}} $$ then $D^{n}u =0$ and for having $n$ arbitrary constants, is a general solution. Thus

$$y = {e^{ax}}\sum\limits_{k = 1}^n {{c_k}{x^{k - 1}}} $$

Thus your expression can be put as $$y = \sum\limits_{k = 0}^n {{a_k}{y^{\left( k \right)}}} = \sum\limits_{k = 0}^n {{a_k}{D^k}y} = \phi \left( D \right)y$$

and you can use the theorem for each root by factoring $\phi(D)$.

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While there are some other good answers to this question, I'll proceed along a different line. Certainly, linear combinations of the solutions $f(x)=p_\lambda(x)e^{\lambda x}$ will work, where $\lambda$ is a root of the characteristic polynomial of the differential equation, $\chi(t) = \sum a_k t^k$, and $\deg p_\lambda$ is less than the multiplicity of $\lambda$ in $\chi$. This we can check by hand, and the uniqueness theorem for solutions to linear differential equations says that the $\mathbb{C}$-vector space spanned by the possible $f(x)$'s exhausts the solutions.

On the other hand, this offers no intuition as to how the solutions were found in the first place. I'll show how to construct them, on the assumption that they represent holomorphic functions near $0$ (this shouldn't be too much of a leap of faith; in any case, it works). With that assumption, let $f(x) = \sum c_n x^n$, and if $f$ solves our equation, we must have $$ a_0 \sum c_n t^n + a_1 \sum n c_nt^{n-1} + \ldots + a_d \sum \frac{n!}{(n-d)!} c_n t^{n-d} = 0.$$ Equating coefficients, it follows that $$a_0c_n + a_1(n+1)c_{n+1}+\ldots + a_d\frac{(n+d)!}{n!}c_{n+d}=0.$$ Next, we introduce $b_n = c_n n!$, such that $$a_0 \frac{b_n}{n!} + a_1 \frac{b_{n+1}}{n!} + \ldots + a_d \frac{b_{n+d}}{n!} = 0.$$ Of course, this implies that $\sum a_k b_{n+k} =0$ for all $n$; that is, the $b_n$ satisfy a linear recurrence. By the general theory of linear recurrences, we have $b_n = \sum p_\lambda(n) \lambda^n$, where $\chi(\lambda)=0$ with multiplicity greater than $\deg p_\lambda$. In turn, we have $c_n = \sum p_\lambda(n) \lambda^n/n!$, so $$f(x) = \sum c_n t^n = \sum_n \sum_\lambda p_\lambda(n)\frac{\lambda^n}{n!}x^n.$$ Here is where the powers of $x$ enter; rewrite the polynomials $p_\lambda$ with respect to the basis $1$, $n$, $n(n-1)$, etc., so the summand corresponding to $\lambda$ reads as $$\sum_n \left(s_0 \frac{\lambda^n}{n!} x^n + s_1n \frac{\lambda^n}{n!}x^n + \ldots s_k \frac{n!}{(n-k)!} \frac{\lambda^n}{n!} x^n\right).$$ Canceling, and recognizing the Taylor series for the exponential, gives us $$s_0 e^{\lambda x} + s_1 (\lambda x) e^{\lambda x} + \ldots + s_k(\lambda x)^k e^{\lambda x}.$$ Now, we just sum the contributions over all $\lambda$ to recognize our solution in the form $$ \sum_\lambda q_\lambda(x) e^{\lambda x}\;,$$ with $q_\lambda(x)$ a polynomial in $\mathbb{C}[x]$ of degree less than the multiplicity of $\lambda$ in $\chi$. By our work, moreover, all holomorphic solutions to this differential equation must be of this form.

Note: with this, and some `backwards reasoning', one can transfer many results between the theory of differential equations with constant coefficients to the theory of linear recurrences. To some extent, they are entirely analogous.

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