数学代写|常微分方程代写ordinary differential equation代考|MATH3331

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• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|常微分方程代写ordinary differential equation代考|Separation of variables

Consider the ODE of the form
$$\frac{d}{d x} y(x)=\frac{f(x)}{g(y(x))} .$$
We assume that $f:\left(a_0, a_1\right) \rightarrow \mathbb{R}$ and $g:\left(b_0, b_1\right) \rightarrow(0, \infty)$ are continuous functions. Wè also assume that there exists $y_0$ in the interval $\left(b_0, b_1\right)$ such that
$$g\left(y_0\right) \neq 0 .$$
We define a function $F:\left(a_0, a_1\right) \times\left(b_0, b_1\right) \rightarrow \mathbb{R}$ by
$$F(x, y)=\int_{y_0}^y g(\xi) d \xi-\int_{x_0}^x f(s) d s, x \in\left(a_0, a_1\right), y \in\left(b_0, b_1\right) .$$
Since $f$ and $g$ are continuous, $F$ is a $C^1$-function. Moreover for every $x_0 \in$ $\left(a_0, a_1\right)$ we have
$$\frac{\partial F}{\partial y}\left(x_0, y_0\right)=g\left(y_0\right) \neq 0 \text {. }$$

Therefore by the implicit function theorem (see Appendix C) there exists $\delta>0$ and a $C^1$-function $\phi:\left(x_0-\delta, x_0+\delta\right) \rightarrow \mathbb{R}$ such that
$$F(x, \phi(x))=\int_{y_0}^{\phi(x)} g(\xi) d \xi-\int_{x_0}^x f(s) d s=F\left(x_0, y_0\right), x \in\left(x_0-\delta, x_0+\delta\right) .$$
One can easily prove that $\phi$ is a solution to (2.1). For, on differentiating (2.3) with respect to $x$ (using the Leibniz rule of differentiation ${ }^1$ ) we get
$$\phi^{\prime}(x) g(\phi(x))-f(x)=0, x \in\left(x_0-\delta, x_0+\delta\right) .$$
This proves that the function $\phi$ which is implicitly given by the relation $F(x, y)=F\left(x_0, y_0\right)$, is a solution to (2.1). In other words, the relation
$$\int^y g(y) d y=\int^x f(x) d x+c, c \in \mathbb{R},$$
where the above integrals are indefinite integrals, defines a solution to (2.1). We now present some examples where this technique is demonstrated.

数学代写|常微分方程代写ordinary differential equation代考|Exact cquations

In this subsection, we present another special form of differential equations called exact equations which can be solved easily. Let $M, N$ be continuous functions in a rectangle
$$R=\left{(x, y):\left|x-x_0\right| \leq a,\left|y-y_0\right| \leq b\right},$$
and $N$ does not vanish in $R$. An ODE of the form
$$N(x, y(x)) y^{\prime}(x)+M(x, y(x))=0,$$
is said to be exact if there exists a $C^1$-function $F: R \rightarrow \mathbb{R}$ such that
$$\frac{\partial F}{\partial x}(x, y)=M(x, y), \quad \frac{\partial F}{\partial y}(x, y)=N(x, y),(x, y) \in R .$$
Example 2.1.8. Show that $y(x) y^{\prime}(x)+x=0$ is an exact equation.
Solution. In order to prove this, we first compare the given equation with (2.18) to get $M(x, y)=x$ and $N(x, y)=y$. It is easy to verify that

$$F(x, y)=\frac{x^2+y^2}{2},$$
satisfies (2.19). Hence the given equation is exact.
We now establish the connection between $F$ and the solutions to (2.18). To this end, we suppose (2.18) is exact and $F$ is known to us. We observe that $\frac{\partial F}{\partial y}=N \neq 0$, in $R$. Let $(\tilde{x}, \tilde{y}) \in \mathbb{R}^2$ satisfy $\left|x_0-\tilde{x}\right|<a$ and $\left|y_0-\tilde{y}\right|<b$. Then by the implicit function theorem there exists an interval $(\tilde{x}-\delta, \tilde{x}+\delta)$, which is denoted by $J$, and a $C^1$-function $\phi: J \rightarrow \mathbb{R}$ such that
$$F(x, \phi(x))=F(\tilde{x}, \tilde{y}), x \in J .$$
Claim. The function $\phi$ is a solution to (2.18).
For, on differentiating (2.20) with respect to $x$ we get
$$\frac{\partial F}{\partial x}(x, \phi(x))+\frac{\partial F}{\partial y}(x, \phi(x)) \phi^{\prime}(x)=0, x \in J .$$
Thus we have
$$M(x, \phi(x))+N(x, \phi(x)) \phi^{\prime}(x)=0, x \in J,$$
which proves that $\phi$ is a solution to (2.18). Hence the claim is proved.
Now, we shall revisit Example 2.1.8 and solve the ODE therein.

常微分方程代写

数学代写|常微分方程代写ordinary differential equation代考|Separation of variables

$$\frac{d}{d x} y(x)=\frac{f(x)}{g(y(x))}$$

$$g\left(y_0\right) \neq 0 .$$

$$F(x, y)=\int_{y_0}^y g(\xi) d \xi-\int_{x_0}^x f(s) d s, x \in\left(a_0, a_1\right), y \in\left(b_0, b_1\right) .$$

$$\frac{\partial F}{\partial y}\left(x_0, y_0\right)=g\left(y_0\right) \neq 0 .$$

$$F(x, \phi(x))=\int_{y_0}^{\phi(x)} g(\xi) d \xi-\int_{x_0}^x f(s) d s=F\left(x_0, y_0\right), x \in\left(x_0-\delta, x_0+\delta\right)$$

$$\phi^{\prime}(x) g(\phi(x))-f(x)=0, x \in\left(x_0-\delta, x_0+\delta\right) .$$

$$\int^y g(y) d y=\int^x f(x) d x+c, c \in \mathbb{R}$$

数学代写|常微分方程代写ordinary differential equation代考|Exact cquations

$$N(x, y(x)) y^{\prime}(x)+M(x, y(x))=0,$$

$$\frac{\partial F}{\partial x}(x, y)=M(x, y), \quad \frac{\partial F}{\partial y}(x, y)=N(x, y),(x, y) \in R .$$

$$F(x, y)=\frac{x^2+y^2}{2}$$

$$F(x, \phi(x))=F(\tilde{x}, \tilde{y}), x \in J .$$

$$\frac{\partial F}{\partial x}(x, \phi(x))+\frac{\partial F}{\partial y}(x, \phi(x)) \phi^{\prime}(x)=0, x \in J .$$

$$M(x, \phi(x))+N(x, \phi(x)) \phi^{\prime}(x)=0, x \in J,$$

有限元方法代写

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MATLAB代写

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