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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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    [Th$i$s](http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-ii-second-order-constant-coefficient-linear-equations/linear-operators-linear-time-invariance/MIT18_03SCF11_s17_6text.pdf) might be helpful to you2012-06-15

2 Answers 2

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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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    Oh, it just called my attention. =)2012-06-16
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The solution of this can go as follows.

THEOREM

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

PROOF

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

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

PROOF

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

Direct calculation

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

Secondly,

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

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

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