Unit D3: Second-Order Linear ODEs: Homogeneous第 D3 单元：二阶线性 ODE：齐次方程

Check that $y_1 = e^{2t}$ and $y_2 = e^{-3t}$ both solve $y'' + y' - 6y = 0$, and confirm that $y = 4y_1 - 5y_2$ solves it too.

For $y_1 = e^{2t}$: $y_1' = 2e^{2t}$, $y_1'' = 4e^{2t}$, so $y_1'' + y_1' - 6y_1 = (4 + 2 - 6)e^{2t} = 0$.

For $y_2 = e^{-3t}$: $y_2' = -3e^{-3t}$, $y_2'' = 9e^{-3t}$, so $y_2'' + y_2' - 6y_2 = (9 - 3 - 6)e^{-3t} = 0$.

By the superposition principle $y = 4y_1 - 5y_2$ is automatically a solution; no separate check is needed.

验证 $y_1 = e^{2t}$，$y_2 = e^{-3t}$ 均满足 $y'' + y' - 6y = 0$，并确认 $y = 4y_1 - 5y_2$ 也满足该方程。

对 $y_1 = e^{2t}$：$y_1' = 2e^{2t}$，$y_1'' = 4e^{2t}$，故 $y_1'' + y_1' - 6y_1 = (4 + 2 - 6)e^{2t} = 0$。

对 $y_2 = e^{-3t}$：$y_2' = -3e^{-3t}$，$y_2'' = 9e^{-3t}$，故 $y_2'' + y_2' - 6y_2 = (9 - 3 - 6)e^{-3t} = 0$。

由叠加原理（superposition），$y = 4y_1 - 5y_2$ 自动是解，无需单独验证。

Superposition is a statement about linear combinations only. With $y_1 = e^{2t}$ and $y_2 = e^{-3t}$ solving $y'' + y' - 6y = 0$, check whether the product $u = y_1 y_2 = e^{-t}$ also solves it.

Compute $u = e^{-t}$, $u' = -e^{-t}$, $u'' = e^{-t}$. Then

So the product is not a solution. This is expected: $e^{-t}$ would be a solution only if $r = -1$ solved $r^2 + r - 6 = 0$, but $(-1)^2 + (-1) - 6 = -6 \ne 0$. Products of solutions live outside the solution space because the operator $L$ is not multiplicative.

叠加原理仅对线性组合成立。以 $y_1 = e^{2t}$，$y_2 = e^{-3t}$ 为 $y'' + y' - 6y = 0$ 的解，检验乘积 $u = y_1 y_2 = e^{-t}$ 是否也是解。

计算 $u = e^{-t}$，$u' = -e^{-t}$，$u'' = e^{-t}$，则

乘积不是解。这是意料之中的：$e^{-t}$ 是解当且仅当 $r = -1$ 满足 $r^2 + r - 6 = 0$，但 $(-1)^2 + (-1) - 6 = -6 \ne 0$。解的乘积不在解空间内，因为算子 $L$ 不具有乘法性。

Given that $y_1 = \cos t$ and $y_2 = \sin t$ both solve $y'' + y = 0$, find the member of the solution family with $y(0) = 3$ and $y'(0) = -2$.

By superposition the general solution is $y = c_1 \cos t + c_2 \sin t$, with $y' = -c_1 \sin t + c_2 \cos t$. Evaluate at $t = 0$:

So $y(t) = 3\cos t - 2\sin t$. The two independent pieces absorb exactly the two pieces of initial data, no more and no fewer, which is the practical meaning of a two-dimensional solution space.

已知 $y_1 = \cos t$ 与 $y_2 = \sin t$ 均满足 $y'' + y = 0$，求满足 $y(0) = 3$，$y'(0) = -2$ 的解。

由叠加原理，通解为 $y = c_1 \cos t + c_2 \sin t$，$y' = -c_1 \sin t + c_2 \cos t$。在 $t = 0$ 代入：

故 $y(t) = 3\cos t - 2\sin t$。两个独立部分恰好吸收两组初始数据，这正是二维解空间的实际含义。

The existence and uniqueness theorem for linear equations guarantees that for any choice of $y(t_0) = \alpha$ and $y'(t_0) = \beta$ there is exactly one solution. Define a map from each solution $y$ to the pair $(y(t_0), y'(t_0))$ in $\mathbb{R}^2$.

This map is linear and, by existence and uniqueness, it is both onto (every pair is achieved) and one to one (a solution is determined by its pair). A linear isomorphism with $\mathbb{R}^2$ means the solution space is itself two dimensional.

Therefore any two linearly independent solutions span all solutions, which is exactly the structure $y = c_1 y_1 + c_2 y_2$ used throughout this unit. To see the linearity of the map explicitly, suppose $y$ and $\tilde y$ are solutions mapping to $(\alpha, \beta)$ and $(\tilde\alpha, \tilde\beta)$. Then $a y + b\tilde y$ is a solution (superposition) whose initial pair is $a(\alpha,\beta) + b(\tilde\alpha,\tilde\beta)$, so the evaluation map respects linear combinations. Injectivity is uniqueness, surjectivity is existence, and a linear bijection preserves dimension.

线性方程的存在唯一定理保证：对任意 $y(t_0) = \alpha$，$y'(t_0) = \beta$，恰好存在唯一解。定义从解 $y$ 到 $\mathbb{R}^2$ 中数对 $(y(t_0), y'(t_0))$ 的映射。

此映射是线性的，且由存在唯一性，它既是满射（每对数值都能达到）也是单射（一个解由其数对唯一确定）。与 $\mathbb{R}^2$ 的线性同构意味着解空间本身是二维的。

因此任意两个线性无关的解张成全部解，这正是本单元通篇使用的结构 $y = c_1 y_1 + c_2 y_2$。

Solve $y'' + y' - 6y = 0$ with $y(0) = 1$, $y'(0) = -8$.

The characteristic equation is $r^2 + r - 6 = (r+3)(r-2) = 0$, so $r = -3$ and $r = 2$. The general solution is

Apply the initial conditions at $t = 0$: $c_1 + c_2 = 1$ and $2c_1 - 3c_2 = -8$. Subtracting twice the first from the second gives $-5c_2 = -10$, so $c_2 = 2$ and $c_1 = -1$.

求解 $y'' - 5y' + 6y = 0$。

特征方程：$r^2 - 5r + 6 = 0$，因式分解得 $(r-2)(r-3) = 0$，根为 $r_1 = 2$，$r_2 = 3$。

通解：$y = c_1 e^{2t} + c_2 e^{3t}$。

Solve $2y'' + 7y' + 3y = 0$ with $y(0) = 4$, $y'(0) = -3$, and describe the long-run behavior.

Characteristic equation: $2r^2 + 7r + 3 = 0$. Factor as $(2r + 1)(r + 3) = 0$, so $r = -\tfrac{1}{2}$ and $r = -3$. Both roots are real, distinct, and negative.

At $t = 0$: $c_1 + c_2 = 4$ and $-\tfrac{1}{2}c_1 - 3c_2 = -3$. Multiply the second by $2$: $-c_1 - 6c_2 = -6$. Add the first: $-5c_2 = -2$, so $c_2 = \tfrac{2}{5}$ and $c_1 = \tfrac{18}{5}$.

Because both exponents are negative, $y(t) \to 0$ as $t \to \infty$. The slower root $-\tfrac{1}{2}$ dominates the tail, so the decay is eventually governed by $e^{-t/2}$. This is the overdamped regime: a monotone return to equilibrium with no oscillation.

求满足 $y'' - 5y' + 6y = 0$，$y(0) = 1$，$y'(0) = 0$ 的解。

由通解 $y = c_1 e^{2t} + c_2 e^{3t}$，$y' = 2c_1 e^{2t} + 3c_2 e^{3t}$。令 $t=0$：

解方程组：$c_2 = -2$，$c_1 = 3$。故 $y = 3e^{2t} - 2e^{3t}$。

Solve $y'' - 3y' = 0$ with $y(0) = 2$, $y'(0) = 6$.

Here $c = 0$, so the characteristic equation is $r^2 - 3r = r(r - 3) = 0$ with roots $r = 0$ and $r = 3$. The root $r = 0$ contributes $e^{0\cdot t} = 1$, a constant solution.

At $t = 0$: $c_1 + c_2 = 2$ and $3 c_2 = 6$, so $c_2 = 2$ and $c_1 = 0$. Thus $y(t) = 2e^{3t}$. The constant mode is present in the general solution even though this particular initial data switches it off.

求解 $y'' + y' - 2y = 0$，$y(0) = 2$，$y'(0) = -3$。

特征方程：$r^2 + r - 2 = 0$，$(r+2)(r-1) = 0$，根为 $r_1 = -2$，$r_2 = 1$。

通解：$y = c_1 e^{-2t} + c_2 e^t$，$y' = -2c_1 e^{-2t} + c_2 e^t$。令 $t=0$：

解得 $c_1 = \tfrac{5}{3}$，$c_2 = \tfrac{1}{3}$。故 $y = \tfrac{5}{3}e^{-2t} + \tfrac{1}{3}e^t$。

Claim: if $r_1 \ne r_2$, then $e^{r_1 t}$ and $e^{r_2 t}$ are linearly independent on any interval. Suppose constants satisfy $c_1 e^{r_1 t} + c_2 e^{r_2 t} = 0$ for all $t$.

Divide by $e^{r_1 t}$ (which is never zero): $c_1 + c_2 e^{(r_2 - r_1)t} = 0$ for all $t$. Differentiate: $c_2 (r_2 - r_1) e^{(r_2 - r_1)t} = 0$. Since $r_2 - r_1 \ne 0$ and the exponential is nonzero, $c_2 = 0$, and then $c_1 = 0$.

The only combination that vanishes identically is the trivial one, so the pair is independent. Equivalently, their Wronskian is $W = (r_2 - r_1)e^{(r_1 + r_2)t} \ne 0$, the test developed in Section 6.

为何恰好是指数函数？若对 $L[y] = 0$ 的解作 Laplace 变换，分母多项式正是 $as^2 + bs + c$。其根将传递函数（transfer function）的极点（poles）决定了系统的模态（modes）$e^{r_k t}$。

此外，算子 $\tfrac{d}{dt}$ 的特征函数（eigenfunctions）正是指数函数：$\tfrac{d}{dt}e^{rt} = r e^{rt}$。因此常系数线性 ODE 的解空间自然由这些特征函数张成，恰如矩阵的特征向量张成对角化基。

The complex roots give formal solutions $e^{(\lambda + i\mu)t}$ and $e^{(\lambda - i\mu)t}$. Factor the real part and apply Euler's formula:

Because the equation has real coefficients, the real and imaginary parts of a complex solution are each real solutions. Taking the real and imaginary parts of $e^{(\lambda + i\mu)t}$ gives

These two functions are linearly independent (one is not a multiple of the other), so their span is the full real solution space, yielding the boxed general solution.

复根情形与简谐振子（simple harmonic oscillator）的物理图像直接对应：$\alpha$ 是衰减率（damping rate），$\beta$ 是阻尼振荡频率（damped frequency）。

若 $\alpha < 0$（欠阻尼，underdamped），解呈振荡衰减；若 $\alpha = 0$（无阻尼），解为纯振荡；若 $\alpha > 0$（不稳定），振幅增长。对实系数二阶 ODE，复根总以共轭对出现，确保通解为实值。

Solve $y'' + 2y' + 5y = 0$ with $y(0) = 1$, $y'(0) = 1$.

Characteristic equation: $r^2 + 2r + 5 = 0$, so $r = \dfrac{-2 \pm \sqrt{4 - 20}}{2} = -1 \pm 2i$. Here $\lambda = -1$ and $\mu = 2$.

At $t = 0$: $y(0) = c_1 = 1$. Differentiate using the product rule: $y'(t) = e^{-t}\big((-c_1 + 2c_2)\cos 2t + (-c_2 - 2c_1)\sin 2t\big)$, so $y'(0) = -c_1 + 2c_2 = 1$. With $c_1 = 1$ this gives $c_2 = 1$.

求解 $y'' + 2y' + 5y = 0$。

特征方程：$r^2 + 2r + 5 = 0$，判别式 $4 - 20 = -16 < 0$，根为 $r = \frac{-2 \pm 4i}{2} = -1 \pm 2i$。

故 $\alpha = -1$，$\beta = 2$，通解：$y = e^{-t}(c_1\cos 2t + c_2\sin 2t)$。

Rewrite the solution $y(t) = e^{-t}(\cos 2t + \sin 2t)$ from Worked Example 3.1 in the single-sinusoid form $y = A e^{-t}\cos(2t - \varphi)$.

Match $c_1 \cos 2t + c_2 \sin 2t = A\cos(2t - \varphi) = A\cos\varphi \cos 2t + A\sin\varphi \sin 2t$. With $c_1 = 1$ and $c_2 = 1$:

Both $\cos\varphi$ and $\sin\varphi$ are positive, so $\varphi$ is in the first quadrant: $\varphi = \pi/4$. Thus

The envelope $\sqrt 2\, e^{-t}$ shows the amplitude decaying, the angular frequency is $2$, and the phase shift is $\pi/4$. The amplitude-phase form makes the physics readable in a way the $c_1, c_2$ form hides.

求满足 $y'' + 4y = 0$，$y(0) = 0$，$y'(0) = 3$ 的解。

特征根 $r = \pm 2i$，故 $\alpha = 0$，$\beta = 2$，通解 $y = c_1\cos 2t + c_2\sin 2t$，$y' = -2c_1\sin 2t + 2c_2\cos 2t$。

令 $t=0$：$c_1 = 0$，$2c_2 = 3$，故 $c_2 = \tfrac{3}{2}$，$y = \tfrac{3}{2}\sin 2t$。

Solve $y'' + 4y = 0$ with $y(0) = 0$, $y'(0) = 6$.

Characteristic equation $r^2 + 4 = 0$ gives $r = \pm 2i$, so $\lambda = 0$, $\mu = 2$, and the solution is an undamped oscillation:

At $t = 0$: $y(0) = c_1 = 0$ and $y'(0) = 2c_2 = 6$, so $c_2 = 3$. Therefore $y(t) = 3\sin 2t$. The zero initial position with nonzero velocity picks out a pure sine, with amplitude $3$ and period $\pi$.

弹簧-质量-阻尼系统满足 $my'' + \gamma y' + ky = 0$。令 $m=1$，$\gamma = 2$，$k = 5$，求解 $y(0) = 1$，$y'(0) = 0$。

特征方程 $r^2 + 2r + 5 = 0$，根 $r = -1 \pm 2i$。通解 $y = e^{-t}(c_1\cos 2t + c_2\sin 2t)$，$y' = e^{-t}[(-c_1+2c_2)\cos 2t + (-c_2-2c_1)\sin 2t]$。

令 $t=0$：$c_1 = 1$，$-c_1 + 2c_2 = 0 \Rightarrow c_2 = \tfrac{1}{2}$。故 $y = e^{-t}(\cos 2t + \tfrac{1}{2}\sin 2t)$。

With a repeated root the equation factors as $a(r - r_0)^2$, so the operator factors as $a\,(D - r_0)^2$, where $D = d/dt$. We seek a second solution of $(D - r_0)^2 y = 0$ beyond $e^{r_0 t}$.

Try $y = u(t)\,e^{r_0 t}$. A short computation gives $(D - r_0)\big(u e^{r_0 t}\big) = u'\,e^{r_0 t}$, so applying the factor twice yields

Setting this to zero forces $u'' = 0$, hence $u = c_1 + c_2 t$. The new piece beyond the constant is $t\,e^{r_0 t}$, exactly the second solution claimed.

重根的出现与临界阻尼（critical damping）直接对应：$b^2 = 4ac$ 时系统处于欠阻尼与过阻尼的边界。物理上，临界阻尼系统返回平衡位置的速度最快而不振荡，解 $(c_1 + c_2 t)e^{rt}$ 的形式——指数衰减乘以一个线性增长——正是这种临界行为的数学体现。

从算子角度，$D^2 - 2rD + r^2 = (D-r)^2$，其核（null space）的维度为 2，由 $e^{rt}$ 与 $t\,e^{rt}$ 张成，类比 Jordan 块（Jordan block）特征向量与广义特征向量的关系。

Solve $y'' - 4y' + 4y = 0$ with $y(0) = 2$, $y'(0) = 5$.

Characteristic equation: $r^2 - 4r + 4 = (r - 2)^2 = 0$, a repeated root $r = 2$. So

At $t = 0$: $y(0) = c_1 = 2$ and $y'(0) = c_2 + 2c_1 = 5$, so $c_2 = 5 - 4 = 1$.

求解 $y'' - 6y' + 9y = 0$。

特征方程：$r^2 - 6r + 9 = (r-3)^2 = 0$，重根 $r = 3$。

通解：$y = (c_1 + c_2 t)e^{3t}$。

Solve $y'' + 6y' + 9y = 0$ with $y(0) = 1$, $y'(0) = -1$, and find whether the solution ever crosses zero for $t > 0$.

Characteristic equation: $r^2 + 6r + 9 = (r + 3)^2 = 0$, a repeated root $r = -3$. So

At $t = 0$: $c_1 = 1$ and $y'(0) = c_2 - 3c_1 = -1$, so $c_2 = 2$. Hence $y(t) = (1 + 2t)e^{-3t}$.

Since $e^{-3t} > 0$, the sign of $y$ is the sign of $1 + 2t$, which is positive for all $t > 0$. So the solution decays to zero without crossing, the hallmark of critical damping: at most one zero crossing, here none.

求满足 $y'' - 6y' + 9y = 0$，$y(0) = 2$，$y'(0) = 1$ 的解。

通解 $y = (c_1 + c_2 t)e^{3t}$，$y' = c_2 e^{3t} + 3(c_1 + c_2 t)e^{3t}$。令 $t=0$：$c_1 = 2$，$c_2 + 3c_1 = 1 \Rightarrow c_2 = 1 - 6 = -5$。

故 $y = (2 - 5t)e^{3t}$。

Confirm by substitution that $y_2 = t e^{3t}$ solves $y'' - 6y' + 9y = 0$ (whose characteristic equation $(r-3)^2 = 0$ has the repeated root $3$).

Differentiate with the product rule: $y_2 = t e^{3t}$, so $y_2' = e^{3t} + 3t e^{3t} = (1 + 3t)e^{3t}$ and $y_2'' = 3e^{3t} + 3(1 + 3t)e^{3t} = (6 + 9t)e^{3t}$.

Substitute:

The bracket collapses to zero, so $t e^{3t}$ is indeed a solution. The general solution is $y = (c_1 + c_2 t)e^{3t}$.

求解 $y'' + 4y' + 4y = 0$，$y(0) = 0$，$y'(0) = 1$。

特征方程：$(r+2)^2 = 0$，重根 $r = -2$。通解 $y = (c_1 + c_2 t)e^{-2t}$。令 $t=0$：$c_1 = 0$，$c_2 - 2c_1 = 1 \Rightarrow c_2 = 1$。

故 $y = t\,e^{-2t}$。

Put $y_2 = v y_1$. Then $y_2' = v' y_1 + v y_1'$ and $y_2'' = v'' y_1 + 2 v' y_1' + v y_1''$. Substitute into $y'' + p y' + q y = 0$ and group by how many times $v$ is differentiated:

The first bracket is zero since $y_1$ solves the equation. With $w = v'$, the rest is a first-order linear equation:

Integrate: $\ln|w| = -2\ln|y_1| - \int p\,dt$, so $w = \dfrac{e^{-\int p\,dt}}{y_1^{\,2}}$. Finally $v = \int w\,dt$ and $y_2 = v y_1$, which is the boxed formula.

降阶法对非常系数方程同样有效，只要能找到一个已知解。例如 Bessel 方程（Bessel equation）和 Legendre 方程（Legendre equation）的第二类解（second kind solutions）均通过对第一类解降阶得到。

更系统地，Abel 恒等式（Abel's identity）$W(t) = W(t_0)\exp\!\left(-\int_{t_0}^t \frac{b(s)}{a(s)}ds\right)$ 给出朗斯基行列式的公式，可以不直接求 $y_2$ 就得到 $W$，再反求 $y_2$。

The equation $t^2 y'' - 3t y' + 4y = 0$ (for $t > 0$) has the solution $y_1 = t^2$. Find a second independent solution.

Write in standard form by dividing by $t^2$: $y'' - \dfrac{3}{t} y' + \dfrac{4}{t^2} y = 0$, so $p(t) = -3/t$. Then $\int p\,dt = -3\ln t$ and $e^{-\int p\,dt} = e^{3\ln t} = t^3$.

So $y_2 = t^2 \ln t$, and the general solution is $y = c_1 t^2 + c_2 t^2 \ln t$.

已知 $y_1 = e^{2t}$ 是 $y'' - 4y' + 4y = 0$ 的解，用降阶法求第二个解。

令 $y = v\,e^{2t}$，则 $y' = (v' + 2v)e^{2t}$，$y'' = (v'' + 4v' + 4v)e^{2t}$。代入：

故 $v'' = 0$，$v = c_1 + c_2 t$。取 $c_1 = 0$，$c_2 = 1$，得 $y_2 = t\,e^{2t}$，与第 4 节重根结论一致。

The constant-coefficient repeated-root case is itself a reduction-of-order problem. Take $y'' - 4y' + 4y = 0$, with known solution $y_1 = e^{2t}$, and recover the second solution by the formula.

In standard form $p(t) = -4$, so $\int p\,dt = -4t$ and $e^{-\int p\,dt} = e^{4t}$. Then

The integral collapses to $\int 1\,dt = t$, so $y_2 = t e^{2t}$, exactly the extra factor of $t$ promised in Section 4. Reduction of order and the factored-operator argument agree.

方程 $t^2 y'' - 3t y' + 4y = 0$（Euler-Cauchy 型）有解 $y_1 = t^2$（可验证）。用降阶法求 $y_2$。

令 $y = v\,t^2$，$y' = v' t^2 + 2vt$，$y'' = v'' t^2 + 4v't + 2v$。代入后消去含 $v$ 的项（因 $y_1=t^2$ 是解）：

令 $w = v'$：$t\,w' + w = 0$，即 $w'/w = -1/t$，解得 $w = 1/t$，$v = \ln t$。故 $y_2 = t^2 \ln t$。

The equation $t^2 y'' - t y' + y = 0$ (for $t > 0$) has the solution $y_1 = t$. Find a second independent solution.

Standard form: divide by $t^2$ to get $y'' - \dfrac{1}{t} y' + \dfrac{1}{t^2} y = 0$, so $p(t) = -1/t$. Then $\int p\,dt = -\ln t$ and $e^{-\int p\,dt} = e^{\ln t} = t$.

So $y_2 = t\ln t$, and the general solution is $y = c_1 t + c_2 t\ln t$. As a check, $y_2' = \ln t + 1$ and $y_2'' = 1/t$, and substituting gives $t^2(1/t) - t(\ln t + 1) + t\ln t = t - t\ln t - t + t\ln t = 0$.

代入 $y = v\,y_1$ 后忘记展开 $y''$，或代入后没有化简含 $y_1$ 满足方程所消去的项。务必完整展开所有导数，再利用 $a y_1'' + b y_1' + c y_1 = 0$ 消去零项，才能得到关于 $w = v'$ 的一阶方程。

（此示例补充说明降阶法的常见问题，方便对比理解。）

Show that $y_1 = e^{2t}$ and $y_2 = e^{-3t}$ form a fundamental set for $y'' + y' - 6y = 0$.

Compute $y_1' = 2e^{2t}$ and $y_2' = -3e^{-3t}$. Then

Since $-5e^{-t} \ne 0$ for all $t$, the two solutions are linearly independent and form a fundamental set, so $y = c_1 e^{2t} + c_2 e^{-3t}$ is the general solution.

计算 $y_1 = e^{2t}$，$y_2 = e^{-3t}$ 的朗斯基行列式，并验证其在 $t=0$ 处不为零。

在 $t=0$ 处 $W = -5 \ne 0$，故两解线性无关，构成基本解组（fundamental set）。

Test whether $y_1 = e^{2t}$ and $y_2 = 3e^{2t}$ form a fundamental set.

Compute $y_1' = 2e^{2t}$ and $y_2' = 6e^{2t}$. Then

The Wronskian is identically zero, so the pair is linearly dependent. Indeed $y_2 = 3 y_1$, a constant multiple, so they span only a one-dimensional space. They cannot serve as a fundamental set, and $c_1 e^{2t} + 3c_2 e^{2t} = (c_1 + 3c_2)e^{2t}$ carries just one effective constant.

检验 $y_1 = e^{t}$，$y_2 = 3e^{t}$ 的朗斯基行列式。

$W \equiv 0$，故 $y_1$ 与 $y_2$ 线性相关（$y_2 = 3y_1$），不构成基本解组，无法给出完整通解。

For $y'' + 2y' + 5y = 0$ (the equation of Worked Example 3.1), use Abel's identity to find the Wronskian up to a constant, then confirm with the solutions $y_1 = e^{-t}\cos 2t$, $y_2 = e^{-t}\sin 2t$.

Here the equation is already in standard form with $p = 2$. Abel gives $W(t) = W(t_0)\exp\!\big(-\int 2\,dt\big) = C e^{-2t}$ for a constant $C$.

Direct computation: $y_1' = e^{-t}(-\cos 2t - 2\sin 2t)$ and $y_2' = e^{-t}(-\sin 2t + 2\cos 2t)$. Then

The bracket simplifies: $-\cos 2t\sin 2t + 2\cos^2 2t + \sin 2t\cos 2t + 2\sin^2 2t = 2$. So $W = 2e^{-2t}$, matching the predicted form $C e^{-2t}$ with $C = 2$, and never zero.

利用 Abel 恒等式验证：对 $y'' + p(t)y' + q(t)y = 0$，朗斯基行列式满足 $W' = -p(t)W$，故 $W(t) = W(0)e^{-\int_0^t p(s)ds}$。若 $p$ 连续，则 $W$ 要么恒为零，要么恒不为零。

对常系数方程 $y'' + by' + cy = 0$，$p = b/a$（常数），$W(t) = W(0)e^{-(b/a)t}$。对例 6.1，$W(0) = -5$，$W(t) = -5e^{-(-3+2)t} = -5e^{-t}$，与直接计算吻合。

Differentiate $W = y_1 y_2' - y_2 y_1'$ using the product rule. The cross terms $y_1' y_2'$ cancel:

Each solution satisfies $y_i'' = -p\,y_i' - q\,y_i$. Substitute:

This is a first-order linear equation $W' = -p\,W$, whose solution is $W(t) = W(t_0)\exp\!\big(-\int_{t_0}^t p\,ds\big)$. The exponential is always positive, so $W$ keeps a fixed sign or is identically zero.

朗斯基行列式与参数变分法（variation of parameters）紧密相关：求非齐次方程 $ay'' + by' + cy = g$ 的特解时，公式为 $y_p = -y_1\int\frac{y_2\,g}{a\,W}\,dt + y_2\int\frac{y_1\,g}{a\,W}\,dt$，其中 $W$ 出现在分母。若 $W = 0$，则公式失效，这再次说明线性无关性（$W \ne 0$）对方法的适用性至关重要。

A mass $m = 1$ hangs on a spring with stiffness $k = 4$ and damping $b$. Classify the motion as $b$ varies, and find the borderline value.

The equation is $y'' + b y' + 4y = 0$ with discriminant $b^2 - 16$. The borderline (critical damping) is $b^2 = 16$, that is $b = 4$.

For $b > 4$ the roots are distinct and negative (overdamped, monotone return). For $b = 4$ the root is repeated, $r = -2$, giving $y = (c_1 + c_2 t)e^{-2t}$ (critical, fastest non-oscillatory return). For $0 < b < 4$ the roots are $-\tfrac{b}{2} \pm i\sqrt{4 - b^2/4}$, an underdamped decaying oscillation.

For example $b = 5$ gives $r^2 + 5r + 4 = (r+1)(r+4) = 0$, roots $-1, -4$: overdamped, $y = c_1 e^{-t} + c_2 e^{-4t}$.

将 $y'' + by' + cy = 0$ 化为一阶系统 $\mathbf{x}' = A\mathbf{x}$，令 $x_1 = y$，$x_2 = y'$，则

$A$ 的特征多项式恰好是 $\lambda^2 + b\lambda + c = 0$，与原 ODE 的特征方程一致。矩阵指数 $e^{At}$ 给出解算子，三类根对应 $A$ 的三类若当形式（Jordan forms）。

Solve $y''' - y'' - y' + y = 0$.

The characteristic equation is $r^3 - r^2 - r + 1 = 0$. Group: $r^2(r - 1) - (r - 1) = (r - 1)(r^2 - 1) = (r - 1)^2(r + 1)$.

So $r = 1$ is a double root and $r = -1$ is simple. The double root contributes $e^{t}$ and $t e^{t}$; the simple root contributes $e^{-t}$.

$n$ 阶常系数线性齐次 ODE $a_n y^{(n)} + \cdots + a_0 y = 0$ 的特征方程为 $a_n r^n + \cdots + a_0 = 0$，有 $n$ 个根（计重数）。每个不同实根 $r$ 贡献 $e^{rt}$，重数为 $m$ 的根贡献 $e^{rt}, te^{rt}, \ldots, t^{m-1}e^{rt}$，复共轭对 $\alpha \pm \beta i$ 贡献 $e^{\alpha t}\cos\beta t$ 与 $e^{\alpha t}\sin\beta t$（类似地处理重数）。通解是这 $n$ 个基本解的线性组合。

Solve $y'''' + 2y'' + y = 0$.

Characteristic equation: $r^4 + 2r^2 + 1 = (r^2 + 1)^2 = 0$. The roots are $r = \pm i$, each of multiplicity $2$. A complex pair $\lambda \pm i\mu$ of multiplicity $2$ contributes the four real solutions $\cos\mu t,\ \sin\mu t,\ t\cos\mu t,\ t\sin\mu t$ (here $\lambda = 0$, $\mu = 1$).

The extra factor of $t$ on the second copy of each function is the same multiplicity rule seen for repeated real roots, applied to the trigonometric family. This is the resonance shape that reappears when forcing in Unit D4.

矩阵指数（matrix exponential）$e^{At} = \sum_{k=0}^\infty \frac{(At)^k}{k!}$ 给出一阶系统 $\mathbf{x}' = A\mathbf{x}$ 的解 $\mathbf{x}(t) = e^{At}\mathbf{x}_0$。计算 $e^{At}$ 时，通过若当分解（Jordan decomposition）$A = PJP^{-1}$，得 $e^{At} = Pe^{Jt}P^{-1}$，其中 $e^{Jt}$ 对角块对应三类根的函数形式。这将整个二阶 ODE 理论纳入线性代数框架。

An RLC circuit obeys $L q'' + R q' + \tfrac{1}{C} q = 0$ with $L = 1$, $C = 1/4$, so $\tfrac{1}{C} = 4$, and resistance $R$. Find the value of $R$ that gives critical damping, and classify $R = 3$.

The equation is $q'' + R q' + 4q = 0$ with discriminant $R^2 - 16$. Critical damping is the borderline $R^2 = 16$, that is $R = 4$ (taking $R > 0$).

For $R = 3$: $R^2 - 16 = 9 - 16 = -7 < 0$, so the roots are complex, $r = -\tfrac{3}{2} \pm i\sqrt{4 - 9/4} = -\tfrac{3}{2} \pm i\tfrac{\sqrt 7}{2}$. The circuit is underdamped: the charge oscillates while decaying, with envelope $e^{-3t/2}$ and angular frequency $\tfrac{\sqrt 7}{2}$.

在处理高阶方程时，将复根对计为两个独立根但只写一个实值解（遗漏 $\sin$ 项或 $\cos$ 项），或将重数为 $m$ 的根只贡献一个基本解而非 $m$ 个。务必根据代数重数写出完整的基本解组。

（此示例说明高阶情形的典型错误，对比理解 7.2 的四个独立解。）