Posted on Categories:Ordinary Differential Equations, 多变量微积分和常微分方程, 常微分方程, 数学代写

# 数学代写|多变量微积分和常微分方程代考Multivariate Calculus & Ordinary Differential Equations代写|MATH340 Exact Equations

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## 数学代写|多变量微积分和常微分方程代考Multivariate Calculus & Ordinary Differential Equations代写|Exact Equations

Let, in the DE of first order and first degree (1.9), the function $f(x, y)=$ $-M(x, y) / N(x, y)$, so that it can be written as
$$M(x, y)+N(x, y) y^{\prime}=0,$$
where $M$ and $N$ are continuous functions having continuous partial derivatives $M_y$ and $N_x$ in the rectangle $S:\left|x-x_0\right|<a,\left|y-y_0\right|<b(0<a, b<$ $\infty)$

Equation (3.1) is said to be exact if there exists a function $u(x, y)$ such that
$$u_x(x, y)=M(x, y) \text { and } u_y(x, y)=N(x, y) .$$
The nomenclature comes from the fact that
$$M+N y^{\prime}=u_x+u_y y^{\prime}$$
is exactly the derivative $d u / d x$.
Once the DE (3.1) is exact its implicit solution is
$$u(x, y)=c .$$
If (3.1) is exact, then from (3.2) we have $u_{x y}=M_y$ and $u_{y x}=N_x$. Since $M_y$ and $N_x$ are continuous, we must have $u_{x y}=u_{y x}$; i.e., for (3.1) to be exact it is necessary that
$$M_y=N_x .$$
Conversely, if $M$ and $N$ satisfy (3.4) then the equation (3.1) is exact. To establish this we shall exhibit a function $u$ satisfying (3.2). We integrate both sides of $u_x=M$ with respect to $x$, to obtain
$$u(x, y)=\int_{x_0}^x M(s, y) d s+g(y) .$$

## 数学代写|多变量微积分和常微分方程代考Multivariate Calculus & Ordinary Differential Equations代写|Elementary First-Order Equations

Suppose in the DE of first order (3.1), M(x,y) $=X_1(x) Y_1(y)$ and $N(x, y)=X_2(x) Y_2(y)$, so that it takes the form
$$X_1(x) Y_1(y)+X_2(x) Y_2(y) y^{\prime}=0 .$$
If $Y_1(y) X_2(x) \neq 0$ for all $(x, y) \in S$, then (4.1) can be written as an exact DE
$$\frac{X_1(x)}{X_2(x)}+\frac{Y_2(y)}{Y_1(y)} y^{\prime}=0$$
in which the variables are separated. Such a DE (4.2) is said to be separable. The solution of this exact equation is given by
$$\int \frac{X_1(x)}{X_2(x)} d x+\int \frac{Y_2(y)}{Y_1(y)} d y=c .$$
Here both the integrals are indefinite and constants of integration have been absorbed in $c$.

Equation (4.3) contains all the solutions of (4.1) for which $Y_1(y) X_2(x) \neq$ 0 . In fact, when we divide (4.1) by $Y_1 X_2$ we might have lost some solutions, and the ones which are not in (4.3) for some $c$ must be coupled with (4.3) to obtain all solutions of (4.1).
Example 4.1. The DE in Example $3.2$ may be written as
$$\frac{1}{x}+\frac{1}{y(1-y)} y^{\prime}=0, \quad x y(1-y) \neq 0$$
for which (4.3) gives the solution $y=(1-c x)^{-1}$. Other possible solutions for which $x\left(y-y^2\right)=0$ are $x=0, y=0$, and $y=1$. However, the solution $y=1$ is already included in $y=(1-c x)^{-1}$ for $c=0$, and $x=0$ is not a solution, and hence all solutions of this DE are given by $y=0, y=(1-c x)^{-1}$.

# 多变量微积分和常微分方程代考

## 数学代写|多变量微积分和常微分方程代考Multivariate Calculus \& Ordinary Differential Equations代写|Exact Equations

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

$$u_x(x, y)=M(x, y) \text { and } u_y(x, y)=N(x, y) .$$

$$M+N y^{\prime}=u_x+u_y y^{\prime}$$

$$u(x, y)=c .$$

$$M_y=N_x .$$

$$u(x, y)=\int_{x_0}^x M(s, y) d s+g(y) .$$

## 数学代写|多变量微积分和常微分方程代考Multivariate Calculus \& Ordinary Differential Equations代写|Elementary First-Order Equations

$$X_1(x) Y_1(y)+X_2(x) Y_2(y) y^{\prime}=0 .$$

$$\frac{X_1(x)}{X_2(x)}+\frac{Y_2(y)}{Y_1(y)} y^{\prime}=0$$

$$\int \frac{X_1(x)}{X_2(x)} d x+\int \frac{Y_2(y)}{Y_1(y)} d y=c .$$

$$\frac{1}{x}+\frac{1}{y(1-y)} y^{\prime}=0, \quad x y(1-y) \neq 0$$
$y=1$ 经包含在 $y=(1-c x)^{-1}$ 为了c $=0$ ，和 $x=0$ 不是解，因此这个 $\mathrm{DE}$ 的所有解都由下式給出 $y=0, y=(1-c x)^{-1}$

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