关于微分方程xy''=y的求解方法咨询
Hey there! Let's work through this differential equation together—you’re totally right that modified Bessel functions are the way to go here, so let’s break this down step by step.
方法一：转化为修正贝塞尔方程
First, let's rewrite the original equation to make it easier to manipulate:
$$xy'' = y \implies y'' - \frac{1}{x}y = 0$$
This is a variable-coefficient second-order linear ODE, so we’ll use a substitution to convert it into a standard modified Bessel equation. Let’s set:
$$t = 2\sqrt{x}$$
First, compute the derivatives of $y$ with respect to $x$ using the chain rule:
- $\frac{dt}{dx} = \frac{1}{\sqrt{x}} = \frac{2}{t}$, so $\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{2}{t} y'_t$
- For the second derivative:
$$\frac{d2y}{dx2} = \frac{d}{dx}\left(\frac{2}{t} y't\right) = \left(-\frac{2}{t^2} y't + \frac{2}{t} y''{tt}\right) \cdot \frac{2}{t} = \frac{4}{t^2} y''{tt} - \frac{4}{t^3} y'_t$$
Substitute this back into the original equation (remember $x = \frac{t^2}{4}$):
$$\frac{t^2}{4} \left(\frac{4}{t^2} y''_{tt} - \frac{4}{t^3} y'_t\right) = y$$
Simplify the left-hand side:
$$y''_{tt} - \frac{1}{t} y'_t = y$$
Multiply through by $t^2$ to get the standard form of a modified Bessel equation:
$$t^2 y''_{tt} - t y'_t - t^2 y = 0$$
To match the classic modified Bessel equation $t^2 z'' + t z' - (t^2 + \nu^2) z = 0$, let’s make one more substitution: $y(t) = t z(t)$. After substituting and simplifying, we end up with:
$$t^2 z'' + t z' - (t^2 + 1) z = 0$$
This is the modified Bessel equation with $\nu = 1$. Its general solution is:
$$z(t) = A I_1(t) + B K_1(t)$$
where $I_1$ is the first-kind modified Bessel function and $K_1$ is the second-kind modified Bessel function.
Substitute back $y(t) = t z(t)$ and $t = 2\sqrt{x}$, and we get the general solution to your original equation:
$$y(x) = 2\sqrt{x} \left( A I_1(2\sqrt{x}) + B K_1(2\sqrt{x}) \right)$$
Here, $A$ and $B$ are arbitrary constants determined by initial/boundary conditions.
方法二：幂级数（弗罗贝尼乌斯方法）
If you want to follow up on your power series attempt, the Frobenius method works here too:
- Assume a solution of the form $y = \sum_{n=0}^\infty a_n x^{n + r}$
- Compute the first and second derivatives, substitute into $xy'' = y$, and equate coefficients of like powers of $x$.
You’ll find the indicial equation gives roots $r = 0$ and $r = 1$. For $r = 1$, you’ll get a valid power series solution that matches the $2\sqrt{x}I_1(2\sqrt{x})$ term in the Bessel function solution. For $r = 0$, since the roots differ by an integer, you’ll need to construct a second linearly independent solution that includes a logarithmic term—this corresponds to the $2\sqrt{x}K_1(2\sqrt{x})$ term, as $K_1$ includes logarithmic components in its series expansion.
备注：内容来源于stack exchange，提问作者CAPA
