常数解解析

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سؤال

解答题目：xy'''=y''y+y'y'; y(0)=-1

الفكرة

Rewriting the differential equation

I’m thinking about substituting $p(y) = y'$ to simplify the third-order ODE. This might make it easier to tackle or provide a clearer method for solving.

Adjusting my approach

Simplifying by setting $v = y'$ and expressing $y''$ as $v'$ , I notice a $y$ term remains. Let me see if another substitution helps here.

Assessing transformation methods

I’m considering known structures, factorization, or simpler forms like $y = \text{constant}$ . The initial condition at $x = 0$ introduces a complication due to the $x$ term in the denominator.

Evaluating potential forms

OK, let me see. I'm working through $y(x) = A x^n$ to avoid singularity at $x = 0$ . Calculating derivatives and matching sides reveals dependencies on $n$ and $A$ .

Verifying the constant

I checked that $y = A$ works by setting $n = 0$ , leading to $y = -1$ as a solution satisfying the initial condition and the differential equation.

Evaluating initial conditions

I'm thinking through whether $y(0) = -1$ and missing $y'(0)$ are enough to pinpoint a unique solution. The absence of full conditions complicates finding a general solution.

Trying another way

I’m noting the ODE's terms and possible substitutions. Explicit relation between terms and derivatives seems to simplify the equation, possibly towards integrating factors or similar methods.

Rearranging terms

I'm working through the ODE by rearranging terms and exploring their derivative relationships. This method might uncover a simpler form or identify patterns that aid in solving the equation.

Redefining variables

I redefined variables to simplify the equation. This new form relates $y''$ to $g(x)$ , potentially offering fresh insights into the ODE's solution.

Revisiting differentials

I’m rethinking the equation by considering the derivative of $x y'$ . This might offer insights into expressing $y$ in terms of $x$ or clarifying the relationship further.

Refining the equation

I’m expressing the equation with $h'(x)$ and $y'(y+2)$ , aiming to link them to $h$ and $y$ for a clearer relationship. This leads me to consider specific forms.

Revisiting strategies

Noticing $y'(y+1)$ , I’m trying a new approach with $z = \frac{y'}{y+1}$ . It’s interesting to see if this leads to a simpler solution.

Rewriting the equation

I redefined variables to simplify the equation with $p$ instead of $y$ and $y'$ . This tweak might help in transforming it into a more manageable form.

Trying another way

I’m working through methods like using $v(x) = \ln p'(x)$ and direct separation of variables to simplify the equation. None are straightforward yet, so I continue exploring.

Evaluating constants

I’m assessing if $C = 0$ simplifies the problem. Using initial conditions, I determined $p(0) = 0$ , which raised some consistency issues with the ordinary differential equation.

Assessing boundary conditions

I’m thinking through boundary conditions to determine $K$ or $C$ , aiming for finite solutions or smoothness at $x = 0$ . This approach holds promise.

Simplifying the equation

I’m working through the equation to find a simpler form for $p$ near $x = 0$ . This approach might reveal a continuous, non-trivial solution.

Establishing relationships

I’m piecing together the power solution for $p$ . The requirement $A (n-1) = A^2 x^n$ suggests $n = 0$ for consistency, meaning $p$ could be zero.

Confirming the solution

I verified that by setting $n = 0$ , $p$ must be zero. This consistency with the original ODE affirmed $y = -1$ , confirming the solution's reliability.

Wondering about unique solutions

I’m considering if $y = -1$ uniquely meets the ODE and initial conditions, given the singularity at $x = 0$ . This leads me to think about other potential solutions.

Confirming the solution

I’m working through $y = -1$ and found it meets the ODE and initial condition $y(0) = -1$ . This suggests $y(x) = -1$ could be the valid solution.

Thought about differential equation solution 持续 39 秒

الإجابة

下面演示一种较为直接的分析方法，说明满足 $y(0)=-1$ 且在 $x=0$ 附近可定义的解只能是常数解 $y(x)\equiv -1$ 。

1. 原方程与初始条件

题目给出的微分方程为

x\,y''' \;=\; y''\,y \;+\;\bigl(y'\bigr)^2,

并且给出初值条件

y(0)\;=\;-1.

我们先检查：若 $y(x)\equiv -1$ ，则

$y'(x)=0,\quad y''(x)=0,\quad y'''(x)=0.$
方程左边： $x\,y'''=x\cdot0=0.$
方程右边： $y''\,y+(y')^2 = 0\cdot(-1)+0=0.$

因此二者相等，为 $0=0$ 。又因 $y(0)=-1$ 与常数解 $y=-1$ 完全吻合，所以

y(x)\equiv -1

显然是原方程的一个解，并且满足题目给出的初始条件。

2. 常数解以外是否可能存在其他解？

这是一个三阶方程，严格来说，要想“唯一”确定解，通常需要给出三个初值(例如 $y(0),\,y'(0),\,y''(0)$ ) 或其他边值条件。题目只给 $y(0)=-1$ 一个条件，但方程在 $x=0$ 处有系数 $x$ 乘着最高阶导数 $y'''$ ，这往往意味着 $x=0$ 是一个奇点(可能是可去奇点，也可能是真奇点)，从而会影响解在 $x=0$ 处的存在与可解析性。

下面简要分析为何在 $x=0$ 附近，如果要保持解的连续可微性(或满足更严格的正则性要求)，只能得到常数解。

2.1 设法寻找一个可积形式

先将原方程改写成

x\,y''' - y''\,y - \bigl(y'\bigr)^2 \;=\;0.

观察到它和下式的导数很接近：

f(x)\;=\;x\,y'' - y'\,y.

因为

f'(x)\;=\;\frac{d}{dx}\bigl(x\,y''\bigr)\;-\;\frac{d}{dx}\bigl(y'\,y\bigr).

其中

\frac{d}{dx}\bigl(x\,y''\bigr) \;=\; y'' \;+\;x\,y''', \quad \frac{d}{dx}\bigl(y'\,y\bigr) \;=\;y''\,y \;+\;\bigl(y'\bigr)^2.

所以

f'(x) \;=\; \bigl(y'' + x\,y'''\bigr)\;-\;\bigl(y''\,y + (y')^2\bigr) \;=\; y'' \;+\;\underbrace{x\,y''' - \bigl[y''\,y + (y')^2\bigr]}_{\text{原方程为0}} \;=\; y''.

这就得到一个简洁的微分方程：

f'(x) \;=\; y''(x).

既然 $f(x)=x\,y'' - y'\,y$ ，则可写成

\frac{d}{dx}\bigl[x\,y'' - y'\,y\bigr] \;=\; y''.

对 $x$ 积分即得

x\,y'' - y'\,y \;=\; \int y''\,dx \;=\; y' + C,

其中 $C$ 是常数。也就是说

x\,y'' \;=\; y'\,y + y' + C \;=\; y'(y+1)\;+\;C.

令

p(x)\;=\;y(x)+1,

则 $p'(x)=y'(x),\;p''(x)=y''(x)$ 。上式就变成

x\,p''(x) \;=\; p'(x)\,p(x) \;+\; C.

若想在 $x=0$ 附近令该方程“正常”成立而无奇异项，一般需要检查在 $x=0$ 处的行为。结合题目的初值 $y(0)=-1$ ，得到

p(0) \;=\; y(0)+1 \;=\; 0.

这样一来，除非再给出 $p'(0)$ 或者其他条件，否则很难保证 $p\neq 0$ 而又在 $x=0$ 保持良好性质。进一步地，若要求在 $x=0$ 无奇点、 $p$ 连续可微，则往往会得到 $C=0$ （这是因为把方程带到 $x=0$ 时，左边是 $0\cdot p''(0)$ 可能要求右边 $p'(0)\,p(0)+C$ 也得是 0，进而导致 $C=0$ ）。这样方程就退化为

x\,p''(x)\;=\;p'(x)\,p(x).

但这在 $p(0)=0$ 附近又往往只能给出平凡解 $p(x)\equiv 0$ ，从而 $y(x)+1\equiv 0$ ，即

y(x)\equiv -1.

结论：当要求在 $x=0$ 连续可定义并且满足 $y(0)=-1$ 时，原方程的解只能是常数解 $y(x)\equiv -1$ 。若不再额外给出关于 $y'(0)$ 、 $y''(0)$ 的边界/初始信息，或者允许在 $x=0$ 处出现奇点(不连续、不可导等)，则很难有其他的“正规”解可选。

3. 最终答案

满足微分方程

x\,y''' = y''\,y + (y')^2

并且满足初值 $y(0)=-1$ 的连续可微解必然是常数解

\boxed{y(x)\equiv -1}.