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I am trying to prove the following statement

Let $P(\lambda)=(\lambda-\lambda_{0})^{r}$where $r$ is a positive integer. Prove that the equation $P(\frac{d}{dt})x(t)=0$ has solutions $t^{i}e^{\lambda_{0}t},i=0,1,\ldots,r-1$

I thought of three ideas that I am having problems to continue with and I need some help,

Idea 1: Induction, this is clear for $r=1$, and maybe since $P'(\lambda)=r(\lambda-\lambda_{0})^{r-1}$ then we could use some inductive step ?

Idea 2: Induction again, maybe use $P(\lambda)=(\lambda-\lambda_{0})(\lambda-\lambda_{0})^{r-1}$ and use some inductive step ?

Idea 3: Use the binomial theorem on $P(\lambda)$ and try do this directly

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This might be helpful to you – Santosh Linkha Jun 15 '12 at 21:32
up vote 2 down vote accepted

The most direct approach is to multiply it out, one factor at a time. (Note my left hand side fixes a typo in the question)

$$\def\ddt{\frac{d}{d t}} \begin{align} \left(\ddt - \lambda_0 \right)^r \left( t^i e^{\lambda_0 t} \right) &= \left(\ddt - \lambda_0 \right)^{r-1} \left(\ddt - \lambda_0 \right) \left( t^i e^{\lambda_0 t} \right) \\&= \left(\ddt - \lambda_0 \right)^{r-1}\left( \ddt\left( t^i e^{\lambda_0 t} \right) - \lambda_0 t^i e^{\lambda_0 t} \right) \\&= \left(\ddt - \lambda_0 \right)^{r-1}\left( i t^{i-1} e^{\lambda_0 t} + \lambda_0 t^i e^{\lambda_0 t} - \lambda_0 t^i e^{\lambda_0 t} \right) \\&= \left(\ddt - \lambda_0 \right)^{r-1}\left( i t^{i-1} e^{\lambda_0 t}\right) \\&= i \left(\ddt - \lambda_0 \right)^{r-1}\left( t^{i-1} e^{\lambda_0 t}\right) \end{align}$$

It should be clear how things work from here.

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Why $\partial t$ and not $\mathrm d t$? – Pedro Tamaroff Jun 16 '12 at 17:13
Probably some combination of habit, and thinking I saw $\partial$ in the OP. Fixed. – Hurkyl Jun 16 '12 at 17:15
Oh, it just called my attention. =) – Pedro Tamaroff Jun 16 '12 at 17:16

The solution of this can go as follows.


If $(D-a)^n y=0$ then $$y=\left(c_1+c_2 x+\cdots +c_n x^{n-1} \right) e^{ax}$$


We begin by the indentity, where $u$ is a function of $x$:

$$\left(D-a \right)(e^{ax} u)=e^{ax} Du$$

We show by induction that

$$\left(D-a \right)^n(e^{ax} u)=e^{ax} D^nu$$

It is true for $n=1$. We show the inductive step:

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

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

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

so by the principle of induction

$$(D-a)^n (e^{ax} u)=e^{ax} D^n u \text{ ; for all } n\in \Bbb N$$

Consider then the equation

$$(D-a)^n y = 0$$

Then set $y=e^{ax} u$. We get that

$$(D-a)^n (e^{ax}u) =e^{ax} D^n u= 0$$

Then $D^n u =0$ from the theorem, so if

$$u=\sum_{k=1}^n c_k x^{k-1}$$ then $D^n u=0$, and since it has $n$ arbitrary cosntants, it is a general solution. Thus

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

Note that each individual term is also a solution.

A more general result is:

THEOREM (Operator shift formula)

Let $$\phi(D) = a_0D^n+a_1 D^{n-1}+\cdots+a_{n-1}D+a_n$$ Then

$$\phi(D) \{e^{mx} F\} =e^{mx} \phi(D+m)\{ F\}$$


Firstly, $$ \tag 1 D \{ e^{mx} F\} = e^{mx} (D+m) F$$

Direct calculation

$$\frac{d}{{dx}}\left( {{e^{mx}}F} \right) = {e^{mx}}\frac{d}{{dx}}F + {e^{mx}}mF = {e^{mx}}\left( {D + m} \right)F$$


$$\tag 2 {D^n}\left\{ {{e^{mx}}F} \right\} = {e^{mx}}{\left( {D + m} \right)^n}F$$

By Lebniz formula

$$ D^n \{e^{mx} F \} = \sum_{k=0}^n {n \choose k} D^k \{e^{mx} \} D^{n-k} \{F \}$$

$$ D^n \{e^{mx} F \} = \sum_{k=0}^n {n \choose k} m^k e^{mx} D^{n-k} \{F \}={e^{mx}}{\left( {D + m} \right)^n}F$$

Finally, since $\phi(D)$ is linear, viz:

$$\eqalign{ & \phi \left( D \right)\left\{ {F + G} \right\} = \phi \left( D \right)\left\{ F \right\} + \phi \left( D \right)\left\{ G \right\} \cr & \phi \left( D \right)\left\{ {k \cdot F} \right\} = k \cdot \phi \left( D \right)\left\{ F \right\} \cr} $$

We get

$$\phi \left( D \right)\left\{ {{e^{mx}}F} \right\} = {e^{mx}}\phi \left( {D + m} \right)\left\{ F \right\}$$

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