## 数学代写|常微分方程代写ordinary differential equation代考|MTH225

2023年1月6日

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## 数学代写|常微分方程代写ordinary differential equation代考|Factorization of differential operators: homogeneous case

In this subsection, we find the solution to the ODE $P(D) y=0$ in terms of the roots of the polynomial $P$. In order to do that, we first recall the following basic results.
Result 1. If all the coefficients of the non-constant polynomial $P$ are real and $z \in \mathbb{C}$ is a root of $P$, then $\bar{z}$ is also a root of $P$. Hence there exist polynomials $Q_1, \ldots, Q_k$ such that degree of each $Q_i, 1 \leq i \leq k$ is either one or two and
$$P(X)=Q_1(X) Q_2(X) \cdots Q_k(X) .$$
Result 2. Let $T_i: V \rightarrow V, 1 \leq i \leq k$ be linear operators defined on a real vector space $V$ and $T_i T_j=T_j T_i, 1 \leq i, j \leq k$. Let $K_i=\left{v \in V: T_i(v)=\right.$ $0}, 1 \leq i \leq k$. Then
$$\left{\sum_{i=1}^k c_i v_i: c_i \in \mathbb{R}, v_i \in K_i\right} \subseteq\left{v \in V: T_1 T_2 \cdots T_k(v)=0\right} .$$
We use these two results to find solutions to the differential equations with constant coefficients. We begin with the following useful result which is motivated by Result 2 .

Lemma 3.1.1. Assume that $I: C(\mathbb{R}) \rightarrow C(\mathbb{R})$ denotes the identity function. Consider the $O D E$
$$\left(D-\alpha_1 I\right)\left(D-\alpha_2 I\right) \cdots\left(D-\alpha_n I\right) y=0, x \in \mathbb{R},$$
where $\alpha_i \in \mathbb{R}$ are distinct real numbers for $1 \leq i \leq n$. Then
$$\phi(x)=c_1 e^{\alpha_1 x}+\cdots+c_n e^{\alpha_n x}, x \in \mathbb{R},$$
where $c_j \in \mathbb{R}, 1 \leq j \leq n$ is a solution to (3.3).
Proof. We first set
$$P(D) y:=\left(D-\alpha_1 I\right)\left(D-\alpha_2 I\right) \cdots\left(D-\alpha_n I\right) y, y \in C^n(\mathbb{R}) .$$
Let $T_i=D-\alpha_i I, 1 \leq i, j \leq n$. Then notice that each $T_j$ is a linear operator and $T_i T_j=T_j T_i, 1 \leq i, j \leq n$. Furthermore, we find that $\phi_j=e^{\alpha_j x}$, is a solution to $\left(D-\alpha_j I\right) \phi_j=0,1 \leq j \leq n$ or $T_j\left(e^{\alpha_j x}\right)=0$. In view of Result 2, it follows that $\phi$ given in (3.4) is a solution to (3.3).

## 数学代写|常微分方程代写ordinary differential equation代考|Factorization of differential operators: non-homogeneous case

In this subsection, we study some methods to find a solution to the following non-homogeneous ODE with constant coefficients
$$\mathscr{L}[y]:=\left(a_n D^n+a_{n-1} D^{n-1}+\cdots+a_0 I\right) y=f,$$
where $a_i \in \mathbb{R}, 0 \leq i \leq n$ and $f$ is a continuous function. Our objective is to find a function in the set $\mathscr{L}^{-1}({f})$. Any member in the set $\mathscr{L}^{-1}({f})$ is called a particular solution to $\mathscr{L}[y]=f$.

We now consider the case where the operator $\mathscr{L}$ is given by
$$\mathscr{L}=\left(D-\alpha_1 I\right)\left(D-\alpha_2 I\right) \cdots\left(D-\alpha_n I\right), \alpha_i \in \mathbb{R} .$$
Since the operator $\mathscr{L}$ is the composition of the operators $D-\alpha_i I, 1 \leq i \leq n$, we readily get that
$$\mathscr{L}^{-1}=\left(D-\alpha_n I\right)^{-1} \cdots\left(D-\alpha_1 I\right)^{-1} .$$
We now find $\left(D-\alpha_1 I\right)^{-1} f$. For, we set $v=\left(D-\alpha_1 I\right)^{-1} f$, then $v$ satisfies
$$\left(D-\alpha_1 I\right) v=f .$$
Equation (3.17) is a first order linear equation and from (2.29) its solution is given by
$$v(x)=e^{\alpha_1 x} \int^x f(x) e^{-\alpha_1 x} d x,$$
where the integral is the indefinite integral. One can apply formula (3.18) récursively to obtain $\left(D-\alpha_n I\right)^{-1} \cdots\left(D-\alpha_1 I\right)^{-1} f$.
Example 3.1.17. Solve $\left(\bar{D}^2-5 \bar{D}+6 \bar{I}\right) y=x, x \in \mathbb{R}$.
Solution: We factorize the given equation as $(D-2 I)(D-3 I) y=x$. By setting $f(x)=x$ and using formula (3.18), we find
$$(D-3 I)^{-1} f=e^{3 x} \int^x x e^{-3 x} d x=-e^{3 x}\left(\frac{x e^{-3 x}}{3}+\frac{e^{-3 x}}{9}\right)=-\frac{x}{3}-\frac{1}{9} .$$
Again using formula (3.18), we compute
$$(D-2 I)^{-1}(D-3 I)^{-1} f=(D-2 I)^{-1}\left(-\frac{x}{3}-\frac{1}{9}\right)=\frac{x}{6}+\frac{5}{36} .$$
Hence
$$\phi(x)=\frac{x}{6}+\frac{5}{36}, x \in \mathbb{R}$$
is a particular solution to the given ODE.

# 常微分方程代写

## 数学代写|常微分方程代写ordinary differential equation代考|Factorization of differential operators: homogeneous case

$$P(X)=Q_1(X) Q_2(X) \cdots Q_k(X) .$$

$$\left(D-\alpha_1 I\right)\left(D-\alpha_2 I\right) \cdots\left(D-\alpha_n I\right) y=0, x \in \mathbb{R}$$

$$\phi(x)=c_1 e^{\alpha_1 x}+\cdots+c_n e^{\alpha_n x}, x \in \mathbb{R},$$

$$P(D) y:=\left(D-\alpha_1 I\right)\left(D-\alpha_2 I\right) \cdots\left(D-\alpha_n I\right) y, y$$

## 数学代写|常微分方程代写ordinary differential equation代考|Factorization of differential operators: non-homogeneous case

$$\mathscr{L}[y]:=\left(a_n D^n+a_{n-1} D^{n-1}+\cdots+a_0 I\right) y=f$$

$$\mathscr{L}=\left(D-\alpha_1 I\right)\left(D-\alpha_2 I\right) \cdots\left(D-\alpha_n I\right), \alpha_i \in \mathbb{R}$$

$$\mathscr{L}^{-1}=\left(D-\alpha_n I\right)^{-1} \cdots\left(D-\alpha_1 I\right)^{-1} .$$

$$\left(D-\alpha_1 I\right) v=f$$

$$v(x)=e^{\alpha_1 x} \int^x f(x) e^{-\alpha_1 x} d x$$

$(D-2 I)(D-3 I) y=x$. 通过设置 $f(x)=x$ 并利 用公式 (3.18)，我们发现
$$(D-3 I)^{-1} f=e^{3 x} \int^x x e^{-3 x} d x=-e^{3 x}\left(\frac{x e^{-3 x}}{3}\right.$$

$$(D-2 I)^{-1}(D-3 I)^{-1} f=(D-2 I)^{-1}\left(-\frac{x}{3}-\frac{1}{9}\right.$$

$$\phi(x)=\frac{x}{6}+\frac{5}{36}, x \in \mathbb{R}$$

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## MATLAB代写

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