Second-Order Differential Equations with Laplace Transforms: IVPs, Step Functions, Impulses, and Convolution

Part I — The Laplace Architecture for Second-Order Equations

The central translation: derivatives become algebra

A differential equation describes how a function changes. That is ideal for modeling but can be awkward for calculation because the unknown \(y\), its derivatives, the forcing input, and the initial conditions appear in different forms. The Laplace transform translates all of them into functions of a new variable \(s\).

In the \(t\)-domain, the equation

is a relationship among derivatives. In the \(s\)-domain it becomes

The unknown is no longer being differentiated. It is multiplied by a quadratic polynomial. The initial conditions have not been postponed or discarded; they appear as concrete numerator terms.

Operator, input, and data

Zero-input and zero-state responses

The master formula naturally separates the solution into two parts:

The first part is called the zero-input response: set the external input to zero but retain the initial state. The second is the zero-state response: set the initial conditions to zero but retain the forcing. This split later becomes the cleanest route to convolution.

The mass-spring-damper model behind the symbols

A standard physical example is

Here \(m\) is mass, \(d\) is damping, \(k\) is stiffness, \(F(t)\) is the applied force, \(y\) is displacement, and \(y^{\prime}\) is velocity. The two initial values describe the displacement and velocity already present when observation begins.

In the transform domain,

The denominator is the system itself. Changing the applied force changes \(F(s)\), but it does not change the system denominator. This is why one transfer function can be reused for a step, a pulse, an impulse, or a completely general input.

What class of problems fits this method best?

The clean algebraic workflow in this lesson is designed for linear constant-coefficient IVPs on \(t\ge0\). Laplace transforms can also handle some variable-coefficient equations and integral equations, but multiplication by \(t\) produces derivatives with respect to \(s\), so the transformed problem may remain differential rather than algebraic. The method is most efficient when the coefficients are constant and the forcing has a known transform or a useful convolution representation.

Part II — Definition, Existence, and the Essential Transform Table

The Laplace transform of a function \(f(t)\), defined for \(t\ge0\), is

for values of \(s\) where the improper integral converges. The factor \(e^{-st}\) suppresses sufficiently moderate growth as \(t\to\infty\). A standard sufficient condition is that \(f\) be piecewise continuous on every finite interval and of exponential order: \(|f(t)|\le Me^{\alpha t}\) for sufficiently large \(t\). These are convenient guarantees, not the most general possible conditions.

Why the one-sided transform fits initial-value problems

The integral begins at \(t=0\), exactly where the initial conditions are supplied. That makes the one-sided Laplace transform a natural tool for causal systems: the input starts at \(t=0\), a switch occurs later at \(t=a\), and an impulse is applied at a specified time.

Essential Laplace transform table

Linearity keeps the workflow manageable

Linearity lets us transform a differential equation term by term. The difficult structural work is concentrated in three places: derivative formulas, algebraic simplification, and inversion.

Four additional transform rules worth knowing

Why matching transforms gives the solution

Under the usual piecewise-continuity and exponential-order hypotheses, the Laplace transform is unique up to equality at isolated points. Once the algebra produces a valid \(Y(s)\), its inverse identifies the causal time-domain solution. That uniqueness is what makes transform tables and partial fractions legitimate solution tools rather than pattern-matching guesses.

Part III — The Derivative Engine and the Master Formula

First derivative

Integration by parts gives

The term (-y(0)) is not an inconvenience. It is the mechanism that inserts the first initial condition into the transformed problem.

Second derivative

Apply the first-derivative formula again:

A second-order equation therefore requires two initial values. In general,

Deriving the master formula

Transform

Using linearity and the derivative identities,

Collect the \(Y\)-terms:

Division yields

How the characteristic equation reappears

The denominator \(as^2+bs+c\) is the characteristic polynomial with \(r\) replaced by \(s\). Its zeros are the poles of the transform. Those poles produce the same exponential modes found by the characteristic-equation method. Laplace transforms do not erase the underlying second-order theory; they package it with the forcing and initial data.

Part IV — The Seven-Step Method

Why the order of the steps matters

Students often know every transform rule but still lose the problem in the algebra. The usual causes are collecting \(Y(s)\) before substituting the initial values, leaving the denominator unfactored, or trying to invert a rational expression before it is decomposed.

The safest workflow is mechanical:

1. Write \(Y(s)=\mathcal L\{y(t)\}\).
2. Transform the equation without mental shortcuts.
3. Replace \(y(0)\) and \(y^{\prime}(0)\) immediately.
4. Move all non-\(Y\) terms to the other side.
5. Factor the full coefficient of \(Y\).
6. Make the expression invertible using partial fractions, shifting, completing the square, or convolution.
7. Check the original problem, not merely the transformed equation.

What to write on an exam

A clean solution should show the derivative transforms, the substituted initial data, the isolated formula for \(Y(s)\), the decomposition, the inverse transform, and the verification. Skipping those transitions may save a few seconds, but it makes sign errors difficult to detect and partial credit difficult to earn.

Part V — Complete Worked IVP from Start to Finish

Solve

Step 1: transform the equation

Let \(Y(s)=\mathcal L\{y(t)\}\). Then

Substitute every transformed piece:

Step 2: solve for \(Y(s)\)

Collect the \(Y\)-terms:

Isolate \(Y\) and factor:

Combine the numerator over the common denominator \(s\):

Step 3: partial fractions

Write

For distinct linear factors, cover each factor and substitute its zero:

Therefore

Step 4: invert term by term

Hence

The terms \(-2e^{-t}\) and \(e^{-2t}\) are transients. Both decay because their exponents are negative. The constant \(2\) is the steady-state response to the constant input \(4\).

Step 5: verify the IVP

First check the initial values:

Now compute

Substitution gives

Fast transform-domain checks

When their hypotheses are satisfied, the initial- and final-value theorems provide fast consistency checks:

For this example, the limits are \(1\) and \(2\). The final-value theorem requires the poles of \(sY(s)\) to lie in the open left half-plane. Do not use it when undamped oscillations or unstable poles prevent a finite final value.

Second complete example: resonant forcing

Consider

The forcing frequency equals the system’s natural frequency. Transforming gives

A standard inverse pair is

Therefore

The factor \(t\) is the signature of resonance. In the transform domain, the forcing contributes another copy of \(s^2+1\), creating a repeated quadratic factor. In the time domain, the oscillation acquires a linearly growing envelope.

Verification is quick: \(y(0)=0\), \(y^{\prime}(0)=0\), and direct differentiation gives \(y^{\prime\prime}+y=\sin t\). This example shows why repeated factors matter even when the denominator has no real linear roots.

Part VI — Partial Fractions and Inverse-Transform Patterns

Solving for \(Y(s)\) is only the midpoint. The inverse transform becomes easy when the expression is rewritten as a sum of recognizable table entries. For rational transforms, that usually means factoring the denominator and using partial fractions.

Choose the decomposition from the denominator

When the cover-up method works

The cover-up shortcut works directly for distinct linear factors. For

the coefficient of \(1/(s-r_1)\) is

Repeated roots and quadratic factors require the full template. The dependable order is: make the fraction proper, factor over the real numbers, write every required term, solve the coefficients, and only then invert.

Complete the square for oscillatory poles

If the denominator contains \(s^2+2\alpha s+\omega_0^2\), write it as

Then split the numerator into a multiple of \(s+\alpha\) plus a constant. That exposes shifted sine and cosine transform pairs and makes the damping factor \(e^{-\alpha t}\) visible.

Part VII — Unit Step Functions and Delayed Forcing

A unit step is an on-switch:

The exact value at the single point \(t=a\) does not affect its Laplace transform or ordinary differential-equation solution away from the switch. What matters is that the input is off before \(a\) and on after \(a\).

Rewrite a piecewise input before transforming

If an input changes from \(f_0(t)\) to \(f_1(t)\) at \(t=a\), write

This says: begin with the old formula, then add exactly the correction needed when the switch turns on.

The second shifting theorem

The most useful delayed form is

Read it in both directions:

The factor \(e^{-as}\) does not change the shape of \(g\). It delays the entire response by \(a\) units and keeps it zero beforehand.

Why the expression must use t minus a

The theorem requires a shifted function \(g(t-a)\), not merely \(u(t-a)g(t)\). If a piecewise formula contains \(u(t-a)q(t)\), introduce \(v=t-a\), rewrite \(q(t)=q(v+a)\), and then transform the resulting function of \(v\). This is the safest way to avoid a wrong exponential factor or a wrong time shift.

First shift versus second shift

A shift inside \(F\), such as \(F(s-a)\), and a separate factor \(e^{-as}\) are different operations. Naming the theorem before using it prevents one of the most common transform-table errors.

Multiple switches fit in one formula

Suppose

Instead of solving three separate IVPs and matching constants, write

Its transform is

Linearity then says the output is a delayed step response beginning at \(t=2\) minus an equal delayed step response beginning at \(t=5\). The system remembers the earlier input, so turning the force off does not generally make the output vanish instantly.

A finite pulse is two switches

A constant pulse of height \(M\) that begins at \(t=a\) and ends at \(t=b\) is

with transform

The first step turns the input on; the second turns the same amount off.

Part VIII — Complete Example with a Switched Input

Solve the initially resting system

1. Transform the equation

Because the initial values are zero, the derivative terms contribute no extra data:

Therefore

2. Decompose the undelayed part

Its inverse transform is

3. Apply the delay

Equivalently,

4. Interpret and verify

The system does nothing before the input turns on. At the switching time, \(g(0)=0\) and \(g^{\prime}(0)=0\), so both \(y\) and \(y^{\prime}\) remain continuous. After the switch, the response approaches the constant steady state \(1/2\), consistent with setting derivatives to zero in \(2y=1\).

Part IX — Dirac Delta Impulses and Jump Behavior

The Dirac delta \(\delta(t-a)\) is not an ordinary finite-valued function. It is an idealized input concentrated at one instant and normalized so that

for every \(\varepsilon>0\). Its defining sampling property is

which immediately gives

Complete impulse-response example

Apply a unit impulse at \(t=1\) to the same initially resting system:

Transforming gives

Therefore

Why the derivative jumps

Integrate the differential equation across a small interval around \(t=1\). The integrals of \(3y^{\prime}\) and \(2y\) vanish in the limit when \(y\) stays bounded, while the impulse contributes 1. Thus

The displacement \(y\) remains continuous, but the velocity changes instantaneously. More generally, for

the ordinary impulse jump is

provided \(y\) itself remains continuous. This is the mathematical version of an instantaneous momentum kick in a mass-spring-damper model.

Step, finite pulse, and impulse are different inputs

The distributional identity \(u^{\prime}(t-a)=\delta(t-a)\) explains why an impulse is the idealized derivative of an instantaneous switch. Correspondingly, integrating an impulse response produces a step response.

Which state variable can jump?

Integrating the ODE across the impulse is safer than memorizing a slogan. In a regular second-order equation with a delta on the right, \(y^{\prime}\) may jump while \(y\) stays continuous. A jump in \(y\) would make \(y^{\prime}\) contain a delta and \(y^{\prime\prime}\) contain a derivative of delta, which is not present in the stated forcing. Different equations or higher distributional inputs can produce different jump structures, so always use the differential equation itself.

Part X — Convolution and the Impulse-Response Formula

Convolution is the rule that combines an arbitrary input with a system’s response to one unit impulse. For causal functions \(f\) and \(g\),

The convolution theorem states

The impulse response

For the zero-initial-condition equation

define the transfer function and impulse response by

Then \(Y=FH\), so

Every infinitesimal piece of the input acts like a tiny delayed impulse. The integral adds all of those delayed impulse responses.

Convolution rebuilds the running example

For \(y^{\prime\prime}+3y^{\prime}+2y=f(t)\),

With constant input \(f(t)=4\), the zero-state response is

The zero-input response determined by \(y(0)=1\) and \(y^{\prime}(0)=0\) is

Adding the two pieces gives

exactly the solution obtained by direct inversion.

The convolution properties that simplify calculations

For causal functions for which the integrals exist, convolution is commutative, associative, and distributive:

The identity element is the impulse:

A delayed impulse shifts a function:

This is the convolution version of the second shifting theorem.

Green’s function and Duhamel’s principle

In this constant-coefficient causal setting, the impulse response \(h\) is also the causal Green’s function for the differential operator. The formula

is a form of Duhamel’s principle: build the response to an extended force by superposing responses to infinitesimal impulses. The vocabulary changes across differential equations, signals, physics, and engineering, but the mathematical object is the same.

Why the convolution integral stops at t

The upper limit is \(t\) because a causal system cannot respond at time \(t\) to an input applied later. Only values \(0\le\tau\le t\) contribute. If an input begins at \(a>0\), the integrand or the unit-step factor automatically removes all contributions before \(a\).

When convolution is the cleanest inverse

Use convolution when \(Y(s)\) is naturally a product \(F(s)H(s)\), the input transform is complicated, or you want a formula that works for an arbitrary forcing function. Partial fractions are usually shorter for a small rational expression. Convolution is more revealing when the system is fixed and the input changes.

Part XI — Transfer Functions, Poles, Transients, and Steady State

For a linear time-invariant system with zero initial conditions, the transfer function is the ratio of output transform to input transform:

The zero-initial-condition requirement matters. Initial data create their own numerator terms, so the full ratio \(Y/F\) is not the transfer function when stored initial energy is present.

One function organizes three important responses

The step response is the accumulated impulse response:

where the derivative relation is interpreted appropriately at discontinuities.

Poles control the natural behavior

The poles are the zeros of \(as^2+bs+c\), after any legitimate cancellations have been considered. Their locations describe the homogeneous modes:

For a proper second-order transfer function with no hidden unstable modes, poles strictly in the open left half-plane give asymptotic decay of the impulse response. Poles on the imaginary axis require separate care because persistent oscillation may prevent a final value.

Transient versus steady-state response

The transient consists of modes that decay as time passes. The steady-state is what remains under a sustained input. In the running example,

The final-value theorem can find that constant without fully inverting:

but only when the poles of \(sY(s)\) are in the open left half-plane. Always check that hypothesis before using the shortcut.

Natural frequency and damping ratio

After dividing by the leading coefficient, a stable second-order denominator is often written

where \(\omega_n>0\) is the undamped natural frequency and \(\zeta\) is the damping ratio. The poles are

This classification can be read before any inverse transform is performed. Laplace algebra gives the exact coefficients; pole geometry gives the qualitative story.

DC gain predicts the constant-input steady state

For a stable system and a constant input \(f(t)=K\), the transform is \(K/s\). When the final-value theorem applies,

The number \(H(0)\) is the DC gain. In the running example, \(H(0)=1/2\), so a constant input of 4 produces steady value 2, while a unit step produces steady value \(1/2\).

A bridge to frequency response

Evaluating a stable transfer function along \(s=i\omega\) describes how the system changes the amplitude and phase of sinusoidal inputs. That is the bridge from an introductory Laplace-transform course to circuits, vibrations, signals, and control theory. The poles still matter: they encode the system’s natural frequencies and damping.

Part XII — Method Selection, Verification, and Common Mistakes

Laplace transforms are powerful because they are systematic, not because they are always shortest. Choose the method that matches the problem.

Which second-order method should you use?

A verification checklist that catches most errors

1. Initial data: Does the formula give the stated \(y(0)\) and \(y^{\prime}(0)\)?
2. ODE: Does substitution reproduce the forcing on each open time interval?
3. Switch timing: Is every delayed term multiplied by \(u(t-a)\) and written in terms of \(t-a\)?
4. Continuity: For a finite step input, are \(y\) and \(y^{\prime}\) consistent across the switch for a regular second-order equation?
5. Impulse jump: Does an impulse of area \(J\) create the required derivative jump \(J/a\)?
6. Long-time behavior: Does the result agree with pole locations and any expected steady state?
7. Transform check: If practical, transform the final answer and compare it with \(Y(s)\).

The twelve most common mistakes

What if the initial time is not zero?

If the data are given at \(t=t_0\), introduce the shifted time \(\tau=t-t_0\) and define \(z(\tau)=y(\tau+t_0)\). Then \(z(0)=y(t_0)\) and \(z^{\prime}(0)=y^{\prime}(t_0)\), so the standard one-sided formulas apply. Remember to rewrite the forcing in terms of \(\tau\) and translate the final answer back to \(t\).

Limits of the simple Laplace workflow

• Nonlinear equations: the transform of a product such as \(y^2\) or \(yy^{\prime}\) does not reduce to a simple product of transforms.
• Variable coefficients: terms such as \(ty^{\prime\prime}\) introduce derivatives of \(Y(s)\), so the transformed problem may still be a differential equation.
• Unknown transform or difficult inversion: a formal expression for \(Y(s)\) may be less useful than a numerical method or a convolution integral.
• Boundary-value problems: the one-sided transform is tailored to data at one initial time, not conditions split across two endpoints.

These are method-selection limits, not failures of the theory. Knowing when Laplace transforms simplify a problem is part of mastering them.

Part XIII — Mastery Map and Practice Set

The complete concept chain

Ten mastery problems

1. Derive \(\mathcal L\{y^{\prime\prime}\}\) from the first-derivative formula.
2. Solve \(y^{\prime\prime}+5y^{\prime}+6y=0\), \(y(0)=2\), \(y^{\prime}(0)=-1\), using Laplace transforms.
3. Solve \(y^{\prime\prime}+y=\sin t\), \(y(0)=0\), \(y^{\prime}(0)=0\).
4. Rewrite the input that equals 0 for \(t<2\), 3 for \(2\le t<5\), and 0 for \(t\ge5\) using unit steps.
5. Find the transform of \(u(t-2)(t-2)^2\).
6. Solve \(y^{\prime\prime}+3y^{\prime}+2y=u(t-2)\) with zero initial data.
7. Find the impulse response of \(y^{\prime\prime}+4y^{\prime}+4y=f(t)\).
8. For \(2y^{\prime\prime}+y=6\delta(t-3)\), determine the jump in \(y^{\prime}\) at \(t=3\).
9. Write the zero-state solution of \(y^{\prime\prime}+3y^{\prime}+2y=f(t)\) as a convolution integral.
10. Explain why the final-value theorem fails for \(Y(s)=1/(s^2+1)\).

Woody mastery check

You understand this chapter when you can reconstruct the derivative formulas, derive the master formula, execute the seven-step chain, translate a piecewise input into steps, interpret \(e^{-as}\), explain the delta jump, and derive \(y_{\mathrm{ZS}}=f*h\) from \(Y=FH\) without looking.