## 数学代写|常微分方程代写ordinary differential equation代考|MATH3303

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|Method of undetermined coefficients

We now present a method to find a solution to the non-homogeneous ODEs (3.16) when the right hand side $f(x)$ is one of the trigonometric functions $\sin (\alpha x)$ or $\cos (\alpha x)$ where $\alpha \in \mathbb{R}$. In this case, we assume that there is a solution is of the form
$$\phi(x)=A \sin (\alpha x)+B \cos (\alpha x),$$
where $A$ and $B$ are unknowns. We substitute this $\phi$ in the given differential equation and obtain a linear system of equations for $A$ and $B$. We solve this system to find $A, B$, which in turn gives us the solution to the given differential equation. This is demonstrated in the following example.

Example 3.1.21. Solve $\left(D^2+D+I\right) y=\sin (3 x), x \in \mathbb{R}$, using the method of undetermined coefficients.

Solution: In the method of undetermined coefficients, we begin with the ansatz
$$\phi(x)=A \cos (3 x)+B \sin (3 x), x \in \mathbb{R} .$$
We need to determine $A$ and $B$ such that $\phi$ is a solution to the given equation. To this end, we compute
\begin{aligned} & \phi^{\prime}(x)=-3 A \sin (3 x)+3 B \cos (3 x), \ & \phi^{\prime \prime}(x)=-9 A \cos (3 x)-9 B \sin (3 x) . \end{aligned}
By substituting the functions $\phi, \phi^{\prime}$, and $\phi^{\prime \prime}$ in the given equation, we get
$$(-8 A+3 B) \cos (3 x)-(3 A+8 B) \sin (3 x)=\sin (3 x) .$$
By comparing the coefficients of $\sin (3 x)$ and $\cos (3 x)$ on both sides of (3.21), we obtain the system
$$-8 A+3 B=0, \quad-3 A-8 B=1 .$$
On solving this system we get that the solution to this system of equations is $A=-\frac{3}{73}$ and $B=-\frac{8}{73}$. Hence the function
$$\phi(x)=-\frac{3}{73} \cos (3 x)-\frac{8}{73} \sin (3 x), x \in \mathbb{R}$$
is a particular solution to the given equation.

## 数学代写|常微分方程代写ordinary differential equation代考|Exponential shift rule

We begin with stating a very useful identity involving the exponential function and a differential operator with constant coefficients. To this end, using the principle of mathematical induction, one can easily prove that for $\alpha \in \mathbb{R}, n \in \mathbb{N}, g \in C^{\infty}(J)$ we have
$$D^n e^{\alpha x} g(x)=e^{\alpha x}(D+\alpha I)^n g(x), x \in J .$$
Therefore if $P(X)$ is any polynomial with real coefficients, then
$$P(D) e^{\alpha x} g(x)=e^{\alpha x} P(D+\alpha I) g(x) .$$
From (3.26) we get the following identity
$$(P(D))^{-1}\left(e^{\alpha x} f(x)\right)=e^{\alpha x}(P(D+\alpha I))^{-1} f(x) .$$
For, put $g=P(D+\alpha I)^{-1} f$ in (3.26) to obtain
$$P(D)\left(e^{\alpha x}(P(D+\alpha I))^{-1} f(x)\right)=e^{\alpha x} f(x)$$
This readily implies identity (3.27). Formally, (3.27) is written as
$$\frac{1}{P(D)} e^{\alpha x} f(x)=e^{\alpha x} \frac{1}{P(D+\alpha I)} f(x)$$
We now consider a special case where $f \equiv 1$. We first recall identity (3.9)
$$P(D) e^{\alpha x}=P(\alpha) e^{\alpha x}, \alpha \in \mathbb{R} .$$
Suppose $\alpha_0 \in \mathbb{R}$ be such that $P\left(\alpha_0\right) \neq 0$, then from (3.9), we obtain
$$(P(D))^{-1} e^{\alpha_0 x}=\frac{e^{\alpha_0 x}}{P\left(\alpha_0\right)} .$$
On the other hand, suppose $P\left(\alpha_0\right)=0$ and $P^{\prime}\left(\alpha_0\right) \neq 0$. Then we nse the technique introduced in Lemma $3.1 .11$ to find a solution to $P(D) y=e^{\alpha_0 x}$. On differentiating (3.9) with respect to $\alpha$ and substituting $\alpha=\alpha_0$, we get
$$P(D) x e^{\alpha_0 x}=P^{\prime}\left(\alpha_0\right) e^{\alpha_0 x}$$

# 常微分方程代写

## 数学代写|常微分方程代写ordinary differential equation代考|Method of undetermined coefficients

$$\phi(x)=A \sin (\alpha x)+B \cos (\alpha x),$$

$$\phi(x)=A \cos (3 x)+B \sin (3 x), x \in \mathbb{R} .$$

$$\phi^{\prime}(x)=-3 A \sin (3 x)+3 B \cos (3 x), \quad \phi^{\prime \prime}(x)$$

$$(-8 A+3 B) \cos (3 x)-(3 A+8 B) \sin (3 x)=\sin (3 x)$$

$$-8 A+3 B=0, \quad-3 A-8 B=1 .$$

$$\phi(x)=-\frac{3}{73} \cos (3 x)-\frac{8}{73} \sin (3 x), x \in \mathbb{R}$$

## 数学代写|常微分方程代写ordinary differential equation代考|Exponential shift rule

$$D^n e^{\alpha x} g(x)=e^{\alpha x}(D+\alpha I)^n g(x), x \in J$$

$$P(D) e^{\alpha x} g(x)=e^{\alpha x} P(D+\alpha I) g(x) .$$

$$(P(D))^{-1}\left(e^{\alpha x} f(x)\right)=e^{\alpha x}(P(D+\alpha I))^{-1} f(x)$$

$$P(D)\left(e^{\alpha x}(P(D+\alpha I))^{-1} f(x)\right)=e^{\alpha x} f(x)$$

(3.9)
$$P(D) e^{\alpha x}=P(\alpha) e^{\alpha x}, \alpha \in \mathbb{R} .$$

$$(P(D))^{-1} e^{\alpha_0 x}=\frac{e^{\alpha_0 x}}{P\left(\alpha_0\right)}$$

$$P(D) x e^{\alpha_0 x}=P^{\prime}\left(\alpha_0\right) e^{\alpha_0 x}$$

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