Part VI · Differential Equations for Dynamic Systems

Chapter 18 — Second-Order Linear ODEs and RLC Responses

Why This Matters

Two energy-storage elements create richer dynamics. Inductors and capacitors can exchange energy, producing oscillation; resistance removes energy, producing damping. Second-order ODEs classify these behaviours through the roots of one quadratic equation.

18.1 Homogeneous constant-coefficient equations

Consider

ay''+by'+cy=0,\qquad a\ne0.

Try $y=e^{st}$ . Since $y'=se^{st}$ and $y''=s^2e^{st}$ ,

(as^2+bs+c)e^{st}=0.

The exponential never equals zero, so $$s$$ must satisfy the characteristic equation

\[as^2+bs+c=0.\]

The nature of its roots determines the solution.

Distinct real roots $$s_1,s_2$$ : $y=C_1e^{s_1t}+C_2e^{s_2t}.$
Repeated real root $$s$$ : $y=(C_1+C_2t)e^{st}.$
Complex roots $\alpha\pm i\beta$ : $y=e^{\alpha t}(C_1\cos\beta t+C_2\sin\beta t).$

Two initial conditions, typically $$y(0)$$ and $$y'(0)$$ , determine $$C_1$$ and $$C_2$$ .

Why exponentials are the trial shape. Differentiating e^{rt} only multiplies it by r, so every derivative in a constant-coefficient equation remains the same shape. Substitution turns the differential equation into an algebraic condition on r; the available roots determine the available response shapes.

y=e^{rt}\quad\Longrightarrow\quad ar^2+br+c=0

Real, repeated, and complex characteristic roots leading to distinct response families.

18.2 RLC damping

For a series RLC natural response, a standard current equation can be written

i''+2\alpha i'+\omega_0^2i=0,

where

\alpha=\frac{R}{2L},\qquad \omega_0=\frac1{\sqrt{LC}}.

The characteristic roots are

s=-\alpha\pm\sqrt{\alpha^2-\omega_0^2}.

If $\alpha>\omega_0$ , roots are distinct real values: the response is over-damped, non-oscillatory. If $\alpha=\omega_0$ , there is a repeated root: critical damping, the fastest non-oscillatory return under this model. If $\alpha<\omega_0$ , roots are complex and the response is under-damped:

i(t)=e^{-\alpha t}\left(C_1\cos\omega_dt+C_2\sin\omega_dt\right), \qquad \omega_d=\sqrt{\omega_0^2-\alpha^2}.

$\omega_d$ is an oscillation frequency only in the under-damped case. It should not be defined with an absolute value to imply oscillation in an over-damped circuit.

Why the discriminant matters. The characteristic roots determine the ingredients available to the response. Two real roots give two decays, a repeated root is the boundary case, and complex conjugate roots combine decay with oscillation.

r^2+2\zeta\omega_n r+\omega_n^2=0

Under-damped, critically damped, and over-damped responses on common axes.

18.3 Non-homogeneous equations

For

\[ay''+by'+cy=q(t),\]

the general solution is

\[y=y_h+y_p,\]

where $$y_h$$ solves the homogeneous equation and $$y_p$$ is one particular solution. For polynomial, exponential, sine, or cosine forcing, the method of undetermined coefficients proposes a trial form whose derivatives have the same family.

If the proposed trial duplicates a term already in $$y_h$$ , multiply the trial by a sufficient power of $$t$$ to make it independent. This resonance correction is essential; otherwise substituted coefficients cannot determine a new solution.

Example — constant forcing

Solve a particular solution for

\[y''+3y'+2y=10.\]

Try $$y_p=A$$ . Its derivatives are zero, so $$2A=10$$ and $$A=5$$ . The homogeneous characteristic equation is $$(s+1)(s+2)=0$$ , hence

y=C_1e^{-t}+C_2e^{-2t}+5.

Initial conditions would determine $$C_1,C_2$$ .

Separate what the system does from what drives it. The homogeneous solution is the natural response allowed by the stored energy and damping. A particular solution accounts for the forcing. Their sum is required because differentiation is linear: substituting each contribution explains a different part of the equation.

\[y=y_h+y_p\]

18.4 Physical interpretation

The homogeneous part contains natural modes determined by stored energy and losses. In a stable damped circuit, it decays and becomes the transient. The particular part is driven by the source and becomes the steady-state response. With $$R=0$$ , the ideal LC equation has undamped sinusoidal solutions; calling it an LC oscillator equation is more precise here than confusing it with the spatial wave-equation PDE.

The RLC circuit is the physical embodiment of a second-order ODE. Each coefficient has a direct electrical correspondence: $$a = L$$ (inductance, opposes current changes), $$b = R$$ (resistance, dissipates energy as heat), and $$c = 1/C$$ (reciprocal capacitance, stores electrostatic energy).

Characteristic roots as system poles

From

Characteristic equation: $L r^2 + R r + \frac{1}{C} = 0$ .
Roots with negative real parts ( $\operatorname{Re}(r) < 0$ ) correspond to stable, decaying physical responses.

Real negative roots ( $$r_1, r_2 < 0$$ ): overdamped — non-oscillatory exponential return.
Repeated real root ( $$r = -R/(2L)$$ ): critically damped — fastest decay without overshoot.
Complex conjugate roots ( $r = -\alpha \pm i\omega_d$ ): underdamped — decaying sinusoidal oscillation at damped frequency $\omega_d = \sqrt{\omega_0^2 - \alpha^2}$ .

Result. The pole positions in the complex $$s$$ -plane completely dictate the transient waveforms and stability of the electrical circuit.

Limitation. Applies to linear time-invariant systems. Non-linear components (such as saturating inductors or semiconductor diodes) require non-linear phase-plane analysis.

18.5 Why the characteristic equation supplies all three cases

Trying $y=e^{st}$ is useful because differentiation only multiplies this trial by $$s$$ or $$s^2$$ . Substitution into a constant-coefficient homogeneous equation therefore reduces a function problem to the algebraic quadratic

\[as^2+bs+c=0.\]

If the roots are distinct, $e^{s_1t}$ and $e^{s_2t}$ are independent solutions, so their linear combination contains two arbitrary constants. If the root is repeated, simply writing the same exponential twice does not give two independent solutions. The second solution is $te^{st}$ , producing

y=(C_1+C_2t)e^{st}.

If roots are $\alpha\pm i\beta$ , Euler's formula turns their exponential solutions into real functions:

e^{(\alpha+i\beta)t}=e^{\alpha t}(\cos\beta t+i\sin\beta t).

Adding and subtracting conjugate solutions gives the real basis $e^{\alpha t}\cos\beta t$ and $e^{\alpha t}\sin\beta t$ . The envelope $e^{\alpha t}$ must remain; it describes growth or decay of the oscillation.

Example — distinct roots and initial conditions

Solve

y''-5y'+6y=0,\qquad y(0)=1,\qquad y'(0)=0.

The characteristic equation factors:

\[s^2-5s+6=(s-2)(s-3)=0.\]

Thus $y=C_1e^{2t}+C_2e^{3t}$ . The first initial condition gives

\[C_1+C_2=1.\]

Differentiate:

y'=2C_1e^{2t}+3C_2e^{3t},

so $$2C_1+3C_2=0$$ at $$t=0$$ . Subtract twice the first equation from the second: $$C_2=-2$$ , hence $$C_1=3$$ . Therefore

y(t)=3e^{2t}-2e^{3t}.

The growth is mathematically correct even though it would signal an unstable physical model if $$y$$ represented a passive circuit state.

Example — repeated root without losing a solution

For $$y''+4y'+4y=0$$ , the characteristic polynomial is

\[s^2+4s+4=(s+2)^2.\]

The correct family is

y=(C_1+C_2t)e^{-2t},

not merely $Ce^{-2t}$ . A second-order ODE needs two independently selectable constants before two independent initial conditions can generally be met.

Example — damped oscillation

Solve $$y''+6y'+25y=0$$ . The roots are

s=\frac{-6\pm\sqrt{36-100}}2=-3\pm4i.

Hence

y=e^{-3t}(C_1\cos4t+C_2\sin4t).

The visible oscillation frequency is $$4$$ rad/s, while $e^{-3t}$ reduces its amplitude. A graph should show crossings continuing while the envelope shrinks, not a sinusoid of fixed height.

[Figure: Three response sketches sharing initial displacement: overdamped non-oscillatory return, critically damped fastest non-oscillatory return, and underdamped decaying oscillation with envelopes

\pm Ae^{-\alpha t}

Why substituting $y = e^{rt}$ produces the characteristic equation. The exponential function is the unique eigenfunction of differentiation: each derivative simply multiplies the function by the constant scalar $$r$$ , converting differential operators into polynomials.

Why the exponential ansatz is complete

Assumptions

The differential equation is linear, homogeneous, with constant coefficients: $$a y'' + b y' + c y = 0$$ .
The solution vector space has dimension 2 (two initial conditions specify a unique trajectory).

$e^{rt}$ satisfies the ODE if and only if $$a r^2 + b r + c = 0$$ .Substituting $y = e^{rt}$ yields $(a r^2 + b r + c)e^{rt} = 0$ . Since $e^{rt} \neq 0$ , $$a r^2 + b r + c = 0$$ must hold.
For distinct roots ( $r_1 \neq r_2$ ), $e^{r_1 t}$ and $e^{r_2 t}$ are linearly independent.The Wronskian determinant $W(e^{r_1 t}, e^{r_2 t}) = (r_2 - r_1)e^{(r_1 + r_2)t} \neq 0$ is non-zero for all $$t$$ .
For a repeated root ( $$r_1 = r_2 = r$$ ), $e^{rt}$ and $t e^{rt}$ are linearly independent.Reduction of order confirms $t e^{rt}$ satisfies the ODE, with non-vanishing Wronskian $W = e^{2rt} \neq 0$ .
For complex roots $\alpha \pm i\beta$ , $e^{\alpha t}\cos(\beta t)$ and $e^{\alpha t}\sin(\beta t)$ form an independent real basis.They are the real and imaginary parts of the complex solution $e^{(\alpha + i\beta)t}$ , both satisfying the real-coefficient ODE.

Conclusion. In all cases, the characteristic quadratic equation produces two linearly independent solutions that span the complete 2D solution space.

Scope. Covers all second-order constant-coefficient linear homogeneous ODEs.

18.6 RLC classification from the discriminant

For the RLC characteristic equation

s^2+2\alpha s+\omega_0^2=0,

the discriminant is $4(\alpha^2-\omega_0^2)$ . Its sign creates the three physical response classes:

Comparison	Roots	Natural response
$\alpha>\omega_0$	distinct real negative for passive components	over-damped, two decays
$\alpha=\omega_0$	repeated negative	critically damped
$\alpha<\omega_0$	$-\alpha\pm i\omega_d$	under-damped, decaying oscillation

Engineering Example — classify before calculating constants

Let $R=4\ \Omega$ , $L=1\ \mathrm H$ , and $C=1/9\ \mathrm F$ . Then

\alpha=\frac{R}{2L}=2\ \mathrm{s}^{-1}, \qquad \omega_0=\frac1{\sqrt{LC}}=3\ \mathrm{rad/s}.

Because $\alpha<\omega_0$ , the response is under-damped and

\omega_d=\sqrt{3^2-2^2}=\sqrt5\ \mathrm{rad/s}.

Only after this classification should one apply initial current and initial derivative conditions to find $$C_1$$ and $$C_2$$ . Calling $\omega_0$ the observed oscillation frequency would be inaccurate when damping is present.

The discriminant classifies physical damping regimes. For $$a y'' + b y' + c y = 0$$ , the discriminant $$D = b^2 - 4ac$$ dictates whether the circuit response is overdamped, critically damped, or underdamped.

Roots and solution forms from the quadratic formula

Starting point. For $$a r^2 + b r + c = 0$$ , the characteristic roots are $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ .

\[D > 0 \implies r_1, r_2 \in \mathbb{R}, \quad y_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}\]
Two distinct real exponential rates: pure decay without oscillation (overdamped).reversible
\[D = 0 \implies r = -\frac{b}{2a}, \quad y_h(t) = (C_1 + C_2 t) e^{-(b/2a)t}\]
Repeated real root: the factor $$t$$ provides the second linearly independent mode (critically damped).reversible
\[D < 0 \implies r = \alpha \pm i\beta, \quad y_h(t) = e^{\alpha t}(C_1 \cos\beta t + C_2 \sin\beta t)\]
Complex conjugate roots with real part $\alpha = -\frac{b}{2a}$ and frequency $\beta = \frac{\sqrt{4ac - b^2}}{2a}$ (underdamped).reversible

Conclusion. The three mathematical cases correspond to three distinct physical dynamics. The sign of $$D = b^2 - 4ac$$ uniquely selects the regime.

18.7 Non-homogeneous equations: choose, test, correct

For $$Ly=q(t)$$ , first solve $$Ly=0$$ to obtain $$y_h$$ . Then choose a trial $$y_p$$ whose derivatives remain in the same family as $$q(t)$$ . Substitute the entire trial and equate coefficients. If the trial overlaps $$y_h$$ , multiply it by $$t$$ once for each overlap order; otherwise it cannot create a genuinely new solution.

Example — polynomial forcing

Solve a particular solution for

\[y''-y'=6t.\]

A degree-one polynomial forcing needs a degree-one trial, but its second derivative may remove terms; use $$y_p=At^2+Bt$$ so that $$y_p''-y_p'$$ can still produce a linear term. Then

y_p'=2At+B,\qquad y_p''=2A,

and substitution gives

\[2A-(2At+B)=6t.\]

Matching the $$t$$ coefficient gives $$-2A=6$$ , so $$A=-3$$ . Matching constants gives $$2A-B=0$$ , so $$B=-6$$ . Thus $$y_p=-3t^2-6t$$ . The homogeneous solution should be added afterwards.

Example — resonance correction, fully checked

Find a particular solution of

y''+y=3\cos t.

Here $y_h=C_1\cos t+C_2\sin t$ , so a usual trial $A\cos t+B\sin t$ overlaps. Multiply by $$t$$ and try $y_p=At\sin t$ . Then

y_p'=A\sin t+At\cos t,

y_p''=2A\cos t-At\sin t.

Therefore

y_p''+y_p=2A\cos t.

Matching $3\cos t$ gives $$A=3/2$$ , hence

y_p=\frac32t\sin t.

The factor $$t$$ explains the increasing amplitude in the ideal undamped resonance model; it is not an arbitrary decoration.

Trial forms for undetermined coefficients. The trial function $$y_p(t)$$ must include all derivatives of the forcing function $$q(t)$$ .

If $q(t) = \text{polynomial of degree } n$ : guess $y_p(t) = A_n t^n + A_{n-1} t^{n-1} + \cdots + A_1 t + A_0$ (complete polynomial).
If $q(t) = C e^{kt}$ : guess $y_p(t) = A e^{kt}$ . If $$k$$ matches a characteristic root, multiply by $$t$$ ; if a repeated root, multiply by $$t^2$$ .
If $q(t) = C\cos(\omega t)$ or $D\sin(\omega t)$ : guess $y_p(t) = A\cos(\omega t) + B\sin(\omega t)$ (always include both sine and cosine terms). Multiply by $$t$$ if $\pm i\omega$ matches a characteristic root (resonance).
If $$q(t)$$ is a sum of terms: guess the sum of the respective individual trial forms (superposition principle).
If $$q(t)$$ is a product (e.g. $t e^{2t}\cos 3t$ ): guess the product of the corresponding trial forms.
After formulating $$y_p$$ , substitute into the ODE, collect like terms, and solve the algebraic system for the unknown coefficients.

18.8 A second-order solution checklist

Put the ODE in standard form and identify whether it is homogeneous.
Form and solve the characteristic equation before writing a solution shape.
Select the correct distinct, repeated, or complex-root family.
If forced, find one particular solution independent of the homogeneous family.
Apply both initial conditions and substitute back into the original equation.
Interpret signs of real parts: negative means decay, zero means persistent oscillation, positive means growth.

Second-order linear ODEs govern oscillatory and resonant systems. The equation $$a y'' + b y' + c y = q(t)$$ models mass-spring-damper systems and RLC circuits.

The characteristic equation

Starting point. Substitute trial solution $y = e^{rt}$ into the homogeneous equation $$a y'' + b y' + c y = 0$$ .

\[y = e^{rt} \implies y' = r e^{rt}, \quad y'' = r^2 e^{rt}\]
Differentiate the exponential ansatz.reversible
\[a(r^2 e^{rt}) + b(r e^{rt}) + c(e^{rt}) = 0\]
Substitute derivatives into the differential equation.reversible
\[(a r^2 + b r + c) e^{rt} = 0\]
Factor out $e^{rt}$ (which is strictly positive and never zero).reversible
\[a r^2 + b r + c = 0\]
The characteristic algebraic quadratic equation must equal zero.reversible

Conclusion. The characteristic equation $$a r^2 + b r + c = 0$$ is obtained by replacing $y'' \to r^2$ , $y' \to r$ , and $y \to 1$ . Its roots $$r_1, r_2$$ define the exponential modes of the system.

Write characteristic equation: $$a r^2 + b r + c = 0$$ .
Compute discriminant: $$D = b^2 - 4ac$$ .
Case 1 ( $$D > 0$$ , distinct real roots $r_1 \neq r_2$ ): $y_h(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}$ (overdamped).
Case 2 ( $$D = 0$$ , repeated real root $$r$$ ): $y_h(t) = (C_1 + C_2 t) e^{rt}$ (critically damped).
Case 3 ( $$D < 0$$ , complex roots $\alpha \pm i\beta$ ): $y_h(t) = e^{\alpha t}(C_1 \cos\beta t + C_2 \sin\beta t)$ (underdamped).
Apply two initial conditions $$y(0) = y_0$$ and $$y'(0) = v_0$$ to find $$C_1$$ and $$C_2$$ .

Common Mistakes

Using the distinct-root formula when the characteristic roots repeat.
Omitting the exponential envelope in an under-damped response.
Calling every non-homogeneous solution a steady state even when the forcing is not persistent or the system is unstable.
Choosing a particular trial that overlaps the homogeneous solution and failing to multiply by $$t$$ .

The general solution to a non-homogeneous ODE is $$y = y_h + y_p$$ . The homogeneous solution $$y_h$$ captures the natural response (determined by the circuit's internal components). The particular solution $$y_p$$ captures the forced steady-state response (driven by the source $$q(t)$$ ).

Method of undetermined coefficients

Starting point. For constant-coefficient ODE $$a y'' + b y' + c y = q(t)$$ , guess a trial form $$y_p(t)$$ matching $$q(t)$$ .

\[q(t) = P_n(t) e^{\gamma t} \implies y_p(t) = t^s Q_n(t) e^{\gamma t}\]
Match the functional form of $$q(t)$$ . The multiplier $$t^s$$ ( $$s = 0, 1, 2$$ ) accounts for overlap with the homogeneous solution, where $$s$$ is the multiplicity of $\gamma$ as a characteristic root.reversible
\[a y_p'' + b y_p' + c y_p = q(t)\]
Substitute $$y_p$$ into the differential equation and equate coefficients of matching terms ( $$t^k$$ , $\cos\omega t$ , $\sin\omega t$ , $e^{kt}$ ) to solve for the unknown constants.reversible

Conclusion. The particular solution mirrors the driving function $$q(t)$$ , modified by $$t^s$$ if there is duplication with the homogeneous natural modes.

1. Solve the homogeneous equation: find $$y_h(t)$$ from the characteristic roots.
2. Inspect the driving term $$q(t)$$ : polynomial, exponential, or sinusoid?
3. Formulate the trial guess $$y_p(t)$$ with undetermined coefficients.
4. If any term in the guess appears in $$y_h(t)$$ , multiply the entire guess by $$t$$ (or $$t^2$$ for double roots).
5. Substitute $$y_p$$ into the ODE and solve the linear system for the unknown coefficients.
6. Write the complete general solution: $$y(t) = y_h(t) + y_p(t)$$ .
7. <strong>Crucial rule:</strong> apply initial conditions to the FULL solution $$y(t) = y_h(t) + y_p(t)$$ , never to $$y_h$$ alone!

Chapter Summary

Second-order constant-coefficient ODEs reduce to a quadratic characteristic equation. Root types determine exponential, repeated-root, or damped-sinusoidal forms. In RLC systems, damping compares $\alpha$ with $\omega_0$ . Forced responses combine a homogeneous transient with a particular driven response.

Exercises

Solve $$y''-5y'+6y=0$$ .
Solve $$y''+4y'+4y=0$$ with $$y(0)=1$$ , $$y'(0)=0$$ .
Classify an RLC response with $\alpha=5\ \mathrm{s}^{-1}$ and $\omega_0=12\ \mathrm{s}^{-1}$ .
Find a particular solution of $y''+y=3\cos t$ , explaining the required trial modification.
Explain, in physical terms, why resistance changes an ideal LC oscillator into a decaying response.
Solve $$y''-5y'+6y=0$$ with $$y(0)=1$$ and $$y'(0)=0$$ .
Solve $$y''+6y'+25y=0$$ and identify its envelope and oscillation frequency.
For $R=4\ \Omega$ , $L=1\ \mathrm H$ , $C=1/9\ \mathrm F$ , classify the natural response and find $\omega_d$ where applicable.
Find a particular solution of $$y''-y'=6t$$ .
Solve for a particular solution of $y''+y=3\cos t$ and explain why a non-resonant trial fails.