柯西问题求助：一阶非线性微分方程解的存在性唯一性及显式求解

Hey there! Let's work through this Cauchy problem together—first tackling the local existence and uniqueness, then diving into the explicit solution you're stuck on.

Proving this is straightforward using the Picard-Lindelöf Theorem (the standard existence-uniqueness theorem for ODEs). Let's frame your equation as $y'(x) = f(x,y)$, where:
$$f(x,y) = \frac{y}{x-1} + x y^2$$
With the initial condition $y(0) = a$.

To apply the theorem, we need two key checks:

• Continuity of $f(x,y)$: In a neighborhood around $(0,a)$, $x \neq 1$ (so the denominator $x-1$ is non-zero), and all terms in $f(x,y)$ are continuous polynomials/monomials. So $f$ is continuous here.
• Local Lipschitz condition in $y$: Compute the partial derivative of $f$ with respect to $y$:
  $$\frac{\partial f}{\partial y} = \frac{1}{x-1} + 2x y$$
  This partial derivative is continuous (and thus bounded) in a small neighborhood around $(0,a)$ (since continuous functions on compact sets are bounded).

Since both conditions are satisfied, the Picard-Lindelöf Theorem guarantees a unique local solution exists around $x=0$.

You mentioned variable separation didn't work—and that makes sense, because this isn't a separable equation. But it is a Bernoulli equation, which follows the standard form:
$$y' + P(x)y = Q(x)y^n$$
Let's rearrange your original equation to match this:
$$y' - \frac{1}{x-1}y = x y^2$$
Here, $n=2$, $P(x) = -\frac{1}{x-1}$, and $Q(x) = x$.

The classic trick for Bernoulli equations is a variable substitution to turn it into a linear ODE. For $n=2$, we use:
$$z = y^{1-n} = y^{-1}$$
Now compute the derivative of $z$ with respect to $x$:
$$z' = -y^{-2} y'$$
Multiply both sides of the rearranged original equation by $-y^{-2}$:
$$-y^{-2}y' + \frac{1}{x-1}y^{-1} = -x$$
Substitute $z$ and $z'$ into this, and we get a first-order linear nonhomogeneous ODE:
$$z' + \frac{1}{x-1}z = -x$$

Solving the Linear ODE with Integrating Factor #

To solve this, we use an integrating factor $\mu(x)$, calculated as:
$$\mu(x) = \exp\left( \int \frac{1}{x-1} dx \right) = |x-1|$$
Since we're looking at solutions near $x=0$, $x-1$ is negative, so we can take $\mu(x) = 1 - x$ (dropping the absolute value sign for this interval).

Multiply both sides of the linear ODE by $\mu(x)$:
$$(1-x)z' + (1-x)\cdot\frac{1}{x-1}z = -x(1-x)$$
Notice the left-hand side is the derivative of $(1-x)z$ (by the product rule):
$$\frac{d}{dx}\left[(1-x)z\right] = -x + x^2$$

Integrate both sides with respect to $x$:
$$(1-x)z = \int (-x + x^2) dx + C = -\frac{1}{2}x^2 + \frac{1}{3}x^3 + C$$

Applying the Initial Condition #

When $x=0$, $y(0)=a$, so $z(0) = \frac{1}{a}$ (we'll handle $a=0$ separately). Plugging $x=0$ into the equation above:
$$(1-0)\cdot\frac{1}{a} = 0 + 0 + C \implies C = \frac{1}{a}$$

Converting Back to $y(x)$ #

Substitute $C$ back in, then replace $z$ with $\frac{1}{y}$:
$$(1-x)\cdot\frac{1}{y} = \frac{1}{a} + \frac{x^3}{3} - \frac{x^2}{2}$$
Solve for $y(x)$:
$$y(x) = \frac{1-x}{\frac{1}{a} + \frac{x^3}{3} - \frac{x^2}{2}}$$
To clean this up, multiply numerator and denominator by $6a$:
$$y(x) = \frac{6a(1-x)}{6 + 2a x^3 - 3a x^2}$$

Special Case: $a=0$ #

If $a=0$, the initial condition is $y(0)=0$. Plugging $y=0$ into the original equation, we see it's a valid solution, and by uniqueness (from the Picard theorem), this is the only solution for $a=0$.

• The denominator of the explicit solution will have roots where the solution blows up, but since we're only concerned with local existence near $x=0$, and the denominator at $x=0$ is $6 \neq 0$, the solution is well-defined in a small neighborhood around $x=0$.
• The key mistake with variable separation was that this isn't a separable equation—recognizing it as a Bernoulli equation is the critical step here!

内容的提问来源于stack exchange，提问作者muserock92