高阶微分方程求解咨询：求y''' + 2y'' + 2y' = 0的通解

Hey there! Let's break down why your solution should satisfy the original equation—odds are you just made a tiny slip-up during differentiation (we've all been there!).

First, let's recap your work to make sure we're on the same page:

• Original ODE: $y''' + 2y'' + 2y' = 0$
• Characteristic equation: $r^3 + 2r^2 + 2r = 0$ → factored to $r(r^2 + 2r + 2) = 0$
• Roots: $r=0$, $r=-1+i$, $r=-1-i$
• General solution: $y = C_1 + C_2e^{-x}\cos(x) + C_3e^{-x}\sin(x)$

Step 1: Compute $y'$ #

Using the product rule for the exponential-trig terms:
$$
\begin{align*}
y' &= 0 + C_2\left(-e^{-x}\cos x - e^{-x}\sin x\right) + C_3\left(-e^{-x}\sin x + e^{-x}\cos x\right) \
&= -C_2e^{-x}(\cos x + \sin x) + C_3e^{-x}(\cos x - \sin x)
\end{align*}
$$

Step 2: Compute $y''$ #

Differentiate $y'$ term by term (again, product rule + chain rule):
$$
\begin{align*}
y'' &= -C_2\left[-e^{-x}(\cos x + \sin x) + e^{-x}(-\sin x + \cos x)\right] + C_3\left[-e^{-x}(\cos x - \sin x) + e^{-x}(-\sin x - \cos x)\right] \
&= -C_2e^{-x}\left(-\cos x - \sin x - \sin x + \cos x\right) + C_3e^{-x}\left(-\cos x + \sin x - \sin x - \cos x\right) \
&= 2C_2e^{-x}\sin x - 2C_3e^{-x}\cos x
\end{align*}
$$

Step 3: Compute $y'''$ #

Differentiate $y''$ one more time:
$$
\begin{align*}
y''' &= 2C_2\left(-e^{-x}\sin x + e^{-x}\cos x\right) - 2C_3\left(-e^{-x}\cos x + e^{-x}(-\sin x)\right) \
&= 2C_2e^{-x}(\cos x - \sin x) + 2C_3e^{-x}(\cos x + \sin x)
\end{align*}
$$

Step 4: Plug into the original ODE #

Now substitute $y'''$, $2y''$, and $2y'$ into $y''' + 2y'' + 2y'$:
$$
\begin{align*}
&2C_2e^{-x}(\cos x - \sin x) + 2C_3e^{-x}(\cos x + \sin x) \
+& 2\left(2C_2e^{-x}\sin x - 2C_3e^{-x}\cos x\right) \
+& 2\left(-C_2e^{-x}(\cos x + \sin x) + C_3e^{-x}(\cos x - \sin x)\right)
\end{align*}
$$

Let's combine like terms for each coefficient:

• For $C_2e^{-x}\cos x$: $2C_2 - 2C_2 = 0$
• For $C_2e^{-x}\sin x$: $-2C_2 + 4C_2 - 2C_2 = 0$
• For $C_3e^{-x}\cos x$: $2C_3 - 4C_3 + 2C_3 = 0$
• For $C_3e^{-x}\sin x$: $2C_3 - 2C_3 = 0$

Everything cancels out to 0, which matches the right-hand side of the original ODE!

The most common mistake here is messing up signs during differentiation (especially with the $e^{-x}$ term, which introduces a negative every time you differentiate it) or mixing up the derivatives of $\cos x$ and $\sin x$. Double-check your differentiation steps—you'll probably spot the tiny error that threw you off.

内容的提问来源于stack exchange，提问作者Stuy