不含因变量y的二阶三阶非线性ODE转化为一阶ODE求解求助

First, let's restate your given ODE clearly:
$$\left(\frac{d^3y}{dx3}\right)^2 + x\frac{d^3y}{dx3} - \frac{d^2y}{dx2} = 0$$

The key insight here is that this equation only involves the second derivative and third derivative of $y$—there's no $y$ or first derivative term. This lets us use a substitution to reduce the order of the ODE to a first-order one, which you can handle with basic techniques.

Step 1: Substitute to Reduce Order #

Let's define a new function $z(x)$ as the second derivative of $y$:
$$z = \frac{d^2y}{dx2}$$
Then, the third derivative of $y$ is just the first derivative of $z$:
$$\frac{d^3y}{dx3} = \frac{dz}{dx}$$

Substitute these into the original ODE, and we get a first-order ODE in terms of $z$ and its derivative:
$$\left(\frac{dz}{dx}\right)^2 + x\frac{dz}{dx} - z = 0$$

We can rearrange this to solve for $z$ directly:
$$z = \left(\frac{dz}{dx}\right)^2 + x\frac{dz}{dx}$$

Step 2: Solve the First-Order ODE #

Let's make one more substitution to simplify this: let $p = \frac{dz}{dx}$ (so $p$ is the third derivative of the original $y$). Now our equation becomes:
$$z = p^2 + x p$$

Now, differentiate both sides with respect to $x$. On the left side, $\frac{dz}{dx} = p$. On the right side, use the product rule:
$$p = 2p\frac{dp}{dx} + p + x\frac{dp}{dx}$$

Simplify this equation by subtracting $p$ from both sides:
$$0 = 2p\frac{dp}{dx} + x\frac{dp}{dx}$$
Factor out $\frac{dp}{dx}$:
$$0 = \frac{dp}{dx}(2p + x)$$

This gives us two separate cases to solve:

Case 1: $\frac{dp}{dx} = 0$ #

If the derivative of $p$ is zero, $p$ is a constant—let's call it $C$. Substitute back into $z = p^2 + x p$:
$$z = C^2 + Cx$$

Remember $z = \frac{d^2y}{dx2}$, so we need to integrate twice to get back to $y$:

1. First integration (to get the first derivative):
   $$\frac{dy}{dx} = \int (C^2 + Cx) dx = C^2x + \frac{C}{2}x^2 + D$$
   (where $D$ is another constant)
2. Second integration (to get $y$):
   $$y = \int \left(C^2x + \frac{C}{2}x^2 + D\right) dx = \frac{C^2}{2}x2 + \frac{C}{6}x^3 + Dx + E$$
   (where $E$ is a third constant)

Case 2: $2p + x = 0$ #

Here, we solve for $p$ directly:
$$p = -\frac{x}{2}$$

Substitute back into $z = p^2 + x p$:
$$z = \left(-\frac{x}{2}\right)^2 + x\left(-\frac{x}{2}\right) = \frac{x^2}{4} - \frac{x^2}{2} = -\frac{x^2}{4}$$

Again, integrate twice to get $y$:

1. First integration:
   $$\frac{dy}{dx} = \int -\frac{x^2}{4} dx = -\frac{x^3}{12} + D$$
2. Second integration:
   $$y = \int \left(-\frac{x^3}{12} + D\right) dx = -\frac{x^4}{48} + Dx + E$$

Final Solutions #

Combining both cases, the general solutions to your original ODE are:

1. The family of cubic polynomials:
   $$y = \frac{C}{6}x^3 + \frac{C^2}{2}x2 + Dx + E$$
2. The quartic solution:
   $$y = -\frac{x^4}{48} + Dx + E$$

Where $C, D, E$ are arbitrary constants.

内容的提问来源于stack exchange，提问作者Tanisha