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# 数学代写|数学分析作业代写Mathematical Analysis代考|Solving First-Order ODEs in Normal Form

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## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Solving First-Order ODEs in Normal Form

In this section and the next we shall examine ODEs of type
$$y^{\prime}=f(x, y),$$
where $f(x, y)$ is a real-valued function. Earlier we saw under which conditions on $f$ equation (4.69) admits a solution satisfying the initial condition $y\left(x_0\right)=y_0$ on the neighbourhood of a point $x_0$.

Hence under the Cauchy theorem’s assumptions, given $x_0$ we determine a solution to (4.69) for any real number $y_0$ (in some interval where $f$ is defined and satisfies the conditions). We then say that the set of solutions depends on a real parameter (called $y_0$ above, but typically denoted with $c$, for constant). Such a solution set is called general integral.

Each element in this family of functions, obtained fixing the parameter $c$, is a particular solution of the differential equation (4.69). Conversely, a solution might not be a particular integral, and these further solutions are called singular integrals. We shall examine, especially in the next section, differential equations admitting singular integrals. Here we will consider special examples of functions $f(x, y)$ for which we are able to compute (more or less explicitly) the general integral of the corresponding equation (4.69).
If the differential equation is given in the form
$$y^{\prime}=f(x) \cdot g(y)$$

with $f(x)$ and $g(y)$ continuous, the equation is called separable (this refers to the fact that the variables are separable, as explained below). Supposing $g(y) \neq 0$ for any $y$, dividing (4.70) by $g(y)$ and integrating in $x$ we obtain
$$\int \frac{y^{\prime}(x)}{g(y(x))} d x=\int f(x) d x$$
This is an expression of the form
$$G(y(x))=F(x)+c$$
where $F$ is a primitive of $f$ and $G$ a primitive of $1 / g$.
When $G$ is invertible, we obtain a family of explicit solutions represented by $y=y(x, c)=G^{-1}(F(x)+c)$

Under the variable change $y=y(x)$ in the left-hand side the integral reduces $(4.70)$ to
$$\int \frac{d y}{g(y)}=\int f(x) d x$$

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Solving First-Order ODEs Not in Normal Form

Consider the differential equation, not in normal form,
$$y=x y^{\prime}+g\left(y^{\prime}\right),$$

where $g$ is differentiable. It is called Clairaut equation. If $y(x)$ is a solution of (4.82) admitting second derivative, by differentiating (4.82) in $x$ we find
$$y^{\prime \prime}\left(x+g^{\prime}\left(y^{\prime}\right)\right)=0,$$
and so either
$$y^{\prime \prime}=0$$
or
$$x+g^{\prime}\left(y^{\prime}\right)=0 .$$
For (4.83) we have $y^{\prime}=c$, and because of (4.82), necessarily
$$y=x c+g(c) .$$
Another solution to (4.82) comes from (4.84). Set $y^{\prime}=t$, so (4.84) and (4.82) give parametric equations
$$\left{\begin{array}{l} x=-g^{\prime}(t) \ y=-t g^{\prime}(t)+g(t) . \end{array}\right.$$
The solution to (4.86) is called a singular integral of the Clairaut equation. It can be proved that this curve is the envelope of the family of lines (4.85), i.e. at any point it is tangent to one line in the family.

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Solving First-Order ODEs in Normal Form

$$y^{\prime}=f(x, y)$$

$$y^{\prime}=f(x) \cdot g(y)$$

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

$$G(y(x))=F(x)+c$$

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

## 数学代写|数学分析代写MATHEMATICAL ANALYSIS代考|Solving First-Order ODEs Not in Normal Form

$$y=x y^{\prime}+g\left(y^{\prime}\right)$$

$$y^{\prime \prime}\left(x+g^{\prime}\left(y^{\prime}\right)\right)=0$$

$$y^{\prime \prime}=0$$

$$x+g^{\prime}\left(y^{\prime}\right)=0$$

$$y=x c+g(c)$$
(4.82) 的另一个解来自 (4.84)。放 $y^{\prime}=t$, 所以 (4.84) 和 (4.82) 给出参数方程 $\$ \$$lleft {$$
x=-g^{\prime}(t) y=-t g^{\prime}(t)+g(t)
$$\正确的。 \ \$$
(4.86) 的解称为 Clairaut 方程的奇异积分。可以证明这条曲线是线族 (4.85) 的包络线，即在任 何一点它都与线族中的一条线相切。

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