Second order constant coefficient equations

A second-order equation determines both a state and its changing rate, so two initial conditions are needed. Constant coefficients allow a precise classification through a quadratic polynomial. Repeated roots and forcing that duplicates a homogeneous mode require modified solution forms rather than division by a vanished denominator.

1 Homogeneous equations and characteristic roots

Consider $y''+ay'+by=0$ with real constants. Substituting $y=e^{rt}$ gives $r^2+ar+b=0$. If roots $r_1,r_2$ are distinct and real, the solution family is

$$y=C_1e^{r_1t}+C_2e^{r_2t}.$$

For a repeated root $r$, write $y=e^{rt}u$. Because $2r+a=0$ and $r^2+ar+b=0$, substitution reduces the equation to $u''=0$, giving $(C_1+C_2t)e^{rt}$. This derives the extra $t$ rather than inserting it as an unexplained rule.

For roots $\alpha\pm i\beta$ with $\beta>0$, the real solutions are $e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t)$. One may verify directly or take real and imaginary parts of the complex exponentials. In each case, the two displayed solutions have independent initial value/derivative pairs. Equivalently their Wronskian is nonzero: it is $(r_2-r_1)e^{(r_1+r_2)t}$ in the distinct-real case, $e^{2rt}$ in the repeated case, and $\beta e^{2\alpha t}$ in the complex case. Thus every prescribed initial pair determines one linear combination. Uniqueness for the equivalent first-order linear system proves these are all solutions.

2 Damped motion

For $y''+2\gamma y'+\omega_0^2y=0$, with $\gamma\ge0$, $\omega_0>0$, the roots are $-\gamma\pm\sqrt{\gamma^2-\omega_0^2}$. When $\gamma<\omega_0$, the solution oscillates with angular frequency $\sqrt{\omega_0^2-\gamma^2}$. Equality gives critical damping; larger $\gamma$ gives two distinct negative real roots.

Worked initial-value problem. For $y''+2y'+5y=0$, $y(0)=1$, $y'(0)=0$, the roots are $-1\pm2i$. The family is $e^{-t}(C_1\cos2t+C_2\sin2t)$. The initial conditions give $C_1=1$ and $-C_1+2C_2=0$, so

$$y=e^{-t}(\cos2t+\tfrac12\sin2t).$$

The decay rate and oscillation frequency are different quantities. A coordinate displacement can change sign even while energy decreases. Indeed $E=(y'^2+\omega_0^2y^2)/2$ satisfies $E'=-2\gamma y'^2\le0$, by multiplying the differential equation by $y'$.

3 Particular solutions and superposition

For $y''+ay'+by=g(t)$, the difference of any two solutions solves the homogeneous equation. Thus one particular solution plus the full homogeneous family gives all solutions. For a constant input with $b\ne0$, a constant trial often works. If $g(t)=P_m(t)e^{\lambda t}$ with real $\lambda$ and a real polynomial $P_m$, use a polynomial of degree $m$ times that exponential, multiplying by $t$ or $t^2$ if $\lambda$ is a simple or double characteristic root. The reason is that substituting $y=e^{\lambda t}u$ shifts each differentiation operator from $D$ to $D+\lambda$; a vanishing constant factor requires one or two integrations in the shifted equation.

For $y''+4y=8$ and zero initial position/velocity, a particular solution is 2. The homogeneous terms enforce the initial data, giving $y=2-2\cos2t$. The particular solution alone does not satisfy those data.

For the resonant equation $y''+4y=\cos2t$, a cosine trial is already homogeneous. A particular solution is $y_p=(t/4)\sin2t$, as two differentiations verify. It also has zero initial position and velocity. The linearly growing envelope is a property of the undamped linear model, not an unlimited-growth prediction for every physical system.

4 A systematic integral construction

For $y''+\omega^2y=g(t)$, $\omega>0$, with zero initial values and continuous $g$, define

$$y(t)=\int_0^t\frac{\sin(\omega(t-s))}{\omega}g(s)ds.$$

Differentiating gives $y'=\int_0^t\cos(\omega(t-s))g(s)ds$; the zero endpoint sine removes the first endpoint term. Differentiating again gives $y''=g(t)-\omega^2y$. Both initial values vanish, so uniqueness proves the formula. This provides a method when a finite undetermined-coefficient trial is inconvenient.

The system uniqueness actually needed for completeness

Let $A:I\to\mathbb R^{d\times d}$ and $b:I\to\mathbb R^d$ be continuous on an interval $I$, and let two differentiable solutions of $X'=A(t)X+b(t)$ have the same value at $t_0\in I$. Their difference $Z$ satisfies $Z'=A(t)Z$, $Z(t_0)=0$. Fix any $t\in I$. On the compact segment joining $t_0$ and $t$, choose $M\ge0$ bounding the operator norm of $A$ and subdivide into finitely many segments of length at most $\delta$ with $M\delta<1$. If $Z(s)=0$ at the beginning of such a segment, integration gives $Z(u)=\int_s^u A(v)Z(v)\,dv$. With $S$ the maximum of $\|Z\|$ on that segment, $S\le M\delta S$, so $S=0$. A finite induction proves $Z(t)=0$. For a segment going backwards from $t_0$, reverse the integral limits before the same estimate. Thus uniqueness holds on all of $I$.

For $y''+ay'+by=g$, use $X=(y,y')^T$, $A=\begin{pmatrix}0&1\\-b&-a\end{pmatrix}$, and inhomogeneous term $(0,g)^T$. The explicit homogeneous pairs above give a solution for every initial pair because their Wronskians are nonzero. Subtracting that solution from any other one and applying the proved system uniqueness establishes completeness of each displayed family. This proof establishes uniqueness for continuous linear systems; it does not claim existence for arbitrary nonlinear systems.

Why the real exponential-polynomial trial works

Here $\lambda$ is real and $P_m$ is a real polynomial of degree $m\ge0$. Setting $y=e^{\lambda t}u$ gives

$$u''+du'+cu=P_m,\qquad d=2\lambda+a,\quad c=\lambda^2+a\lambda+b.$$

If $c\ne0$, substitute $u=\sum_{j=0}^m q_jt^j$. Descending from the coefficient of $t^m$, each new equation has the nonzero coefficient $c$ multiplying the next unknown $q_j$; derivatives use only higher coefficients already determined. Hence there is a unique polynomial trial of degree at most $m$. If $c=0$, $d\ne0$, first solve $v'+dv=P_m$ by the same descending coefficient argument. Integrate $u'=v$ with constant term zero; then $u=tQ_m$ for a polynomial of degree at most $m$. If $c=d=0$, twice integrate $u''=P_m$ with both integration constants zero, giving $u=t^2Q_m$. These are exactly multiplicities zero, one and two of the real root $\lambda$. The trigonometric forcing examples are verified directly by differentiation; no theorem about complex series is being presumed here.

5 Graded practice

E1 Foundation. Solve $y''-3y'+2y=0$ with $y(0)=0$, $y'(0)=1$.

E2 Repeated root. Solve $y''+6y'+9y=0$ with $y(0)=2$, $y'(0)=-3$.

E3 Forcing. Solve $y''+y=3$ with $y(0)=1$, $y'(0)=2$.

E4 Resonance check. Verify that $y=(t/4)\sin2t$ solves $y''+4y=\cos2t$ and the two zero initial conditions.

6 Full solutions

S1. Roots are 1 and 2. Write $y=C_1e^t+C_2e^{2t}$. Conditions give $C_1+C_2=0$, $C_1+2C_2=1$, hence $y=e^{2t}-e^t$.

S2. The repeated root is $-3$. Write $y=(C_1+C_2t)e^{-3t}$. Conditions give $C_1=2$ and $C_2-3C_1=-3$, so $C_2=3$ and $y=(2+3t)e^{-3t}$.

S3. Use particular solution 3 and homogeneous family $C_1\cos t+C_2\sin t$. Initial conditions give $C_1=-2$, $C_2=2$, so $y=3-2\cos t+2\sin t$.

S4. $y'=\sin2t/4+t\cos2t/2$ and $y''=\cos2t-t\sin2t$. Adding $4y=t\sin2t$ leaves $\cos2t$. Both $y(0)$ and $y'(0)$ are zero, confirming the full initial-value solution.

Summary

Characteristic roots classify homogeneous constant-coefficient equations, including the repeated-root case. A particular solution must be combined with homogeneous terms to satisfy initial data. Resonant forcing overlaps a natural mode and changes the trial form; an integral response formula supplies a systematic alternative.