关于二阶线性齐次常系数微分方程通解形式来源的技术问询

Hey there! Let me walk through this with you since you already have a solid start on the distinct real roots case, and I can help unpack the other two scenarios clearly.

先确认你对实根情况的理解完全正确！ #

Your reasoning for the distinct real roots case is totally on point. Let me recap it briefly to align:
We rewrite the operator equation $(D^2 + aD + b)y = 0$ as $(D - r_0)(D - r_1)y = 0$ when the characteristic equation has distinct real roots $r_0, r_1$. Each factor gives a straightforward first-order ODE:

• $(D - r_0)y = 0$ translates to $y' = r_0 y$, whose solution is $y = c_0 e^{r_0 x}$
• $(D - r_1)y = 0$ translates to $y' = r_1 y$, whose solution is $y = c_1 e^{r_1 x}$
  Since the solution space of linear homogeneous ODEs is a linear space, we can combine these solutions linearly to get the general form $y = c_0 e^{r_0 x} + c_1 e^{r_1 x}$. Great job getting that part right!

再来说重复实根的情况 #

Suppose the characteristic equation has a single repeated root $r$ (so our operator equation becomes $(D - r)^2 y = 0$). We already know one solution: $y_1 = e^{r x}$, since applying $(D - r)$ to it gives zero. But since this is a second-order ODE, we need a second linearly independent solution to span the 2-dimensional solution space.

Here's a standard trick called reduction of order: assume the second solution has the form $y_2 = v(x) e^{r x}$, where $v(x)$ is some unknown function we need to solve for. Let's compute its derivatives and plug into the ODE:

1. First derivative: $y_2' = v' e^{r x} + r v e^{r x} = e^{r x}(v' + r v)$
2. Second derivative: $y_2'' = e^{r x}(v'' + r v') + r e^{r x}(v' + r v) = e^{r x}(v'' + 2r v' + r^2 v)$

Now substitute into $(D - r)^2 y = y'' - 2r y' + r^2 y = 0$:
$$
e^{r x}(v'' + 2r v' + r^2 v) - 2r e^{r x}(v' + r v) + r^2 e^{r x} v = 0
$$
Factor out $e^{r x}$ (which is never zero, so we can safely divide both sides by it):
$$
v'' + 2r v' + r^2 v - 2r v' - 2r^2 v + r^2 v = 0
$$
Simplify all the terms, and everything cancels out except $v'' = 0$. Integrate twice: $v' = c_1$, $v = c_1 x + c_2$. We can pick $c_1 = 1$, $c_2 = 0$ to get a simple linearly independent solution, so $y_2 = x e^{r x}$. That's why the general solution for repeated roots is $y = c_1 e^{r x} + c_2 x e^{r x}$!

最后是复共轭根的情况 #

Suppose the characteristic equation has complex conjugate roots $z_1 = \alpha + \beta i$ and $z_2 = \alpha - \beta i$. Using the same logic as distinct real roots, the complex-valued general solution is:
$$
y = c_1 e^{(\alpha + \beta i)x} + c_2 e^{(\alpha - \beta i)x}
$$
But in most practical contexts, we want real-valued solutions. Let's use Euler's formula here: $e^{i\theta} = \cos\theta + i\sin\theta$. Expand the complex solutions:

• $e^{(\alpha + \beta i)x} = e^{\alpha x} e^{i\beta x} = e^{\alpha x}(\cos(\beta x) + i\sin(\beta x))$
• $e^{(\alpha - \beta i)x} = e^{\alpha x} e^{-i\beta x} = e^{\alpha x}(\cos(\beta x) - i\sin(\beta x))$

Since the solution space is linear, any linear combination of these solutions is also a solution. Let's construct real-valued combinations:

• Let $y_1 = \frac{1}{2}(e^{(\alpha + \beta i)x} + e^{(\alpha - \beta i)x}) = e^{\alpha x}\cos(\beta x)$
• Let $y_2 = \frac{1}{2i}(e^{(\alpha + \beta i)x} - e^{(\alpha - \beta i)x}) = e^{\alpha x}\sin(\beta x)$

These are two linearly independent real-valued solutions, so the general real solution simplifies to:
$$
y = e^{\alpha x}(c_1 \cos(\beta x) + c_2 \sin(\beta x))
$$

备注：内容来源于stack exchange，提问作者stats_b