Laplace Transform Derivatives, Integrals & Solving ODEs

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Parts 1 and 2 were setup. This is the part where Laplace Transform stops being abstract and starts solving real problems. The moment you apply the derivative transform formula, a differential equation becomes an algebraic equation in $Y(s)$ . Solve for $Y(s)$ using ordinary algebra, then invert using the partial fractions from Part 2, and the ODE is done — no homogeneous solution, no particular solution, no undetermined coefficients. This post covers the derivative transform, the integration transform, the second shifting theorem, and solving initial value problems, with 10 fully worked problems at exam-level depth. This is Part 3 of the Laplace Transform Series.

📋 BOARD EXAM RELEVANCE

EE (Electrical Engineer) — Very high frequency. Every transient circuit analysis problem eventually requires solving a second-order ODE, and the Laplace method is consistently the fastest path on the boards.
ECE (Electronics Engineer) — Very high frequency. Control systems questions that ask for system response to a given input are solved directly by the four-step IVP method in this post.
ME (Mechanical Engineer) — High frequency for this board specifically. Vibration analysis — natural frequency, damping ratio, steady-state displacement — all require solving second-order ODEs of exactly the form covered here.
CE (Civil Engineer) — Low frequency. Structural dynamics electives reference this method but it rarely appears as a standalone item on the core CE board.
ChE, GeE, MetE, MinE, Naval Architect and Marine Engineer — Low to rare. Process control and some instrumentation topics use Laplace-based ODE solutions, but these are not core items for most of these boards.

Bottom line: If you take EE, ECE, or ME boards, the four-step IVP pattern in this post is one of the highest-value skills in the entire series. Combine it with Part 2’s decomposition methods and you can solve virtually any Laplace-based board problem in one sitting.

Transform of Derivatives

Every time a function gets differentiated, its Laplace Transform trades that derivative for multiplication by $s$ plus a subtracted initial condition. This is the core exchange that makes the method work.

FIRST AND SECOND DERIVATIVE FORMULAS

$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$

$\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0)$

Each differentiation order brings down one more factor of $s$ and subtracts one more initial condition evaluated at $t = 0$ . The number of initial conditions needed always matches the order of the ODE.

Transform of an Integral

INTEGRATION PROPERTY

$\mathcal{L}\left\{\int_0^t f(\tau)\,d\tau\right\} = \dfrac{F(s)}{s}$

Integration in the time domain divides by $s$ in the $s$ -domain. This is the mirror of differentiation, which multiplies by $s$ . The integration property appears in circuit problems where a capacitor voltage is expressed as an integral of current.

The Second Shifting Theorem — Unit Step Functions

SECOND SHIFTING THEOREM

$\mathcal{L}\{f(t-a)\,u(t-a)\} = e^{-as}F(s)$

The unit step function $u(t-a)$ switches a function on at $t = a$ — it is zero for $t < a$ and one for $t \geq a$ . This is how board problems model a switch closing, a voltage turning on, or a force being applied suddenly at a specific time. The transform of any time-shifted, step-gated function is the original transform multiplied by $e^{-as}$ .

The Four-Step IVP Pattern

Every initial value problem on the boards follows the same four steps regardless of how the equation looks on the page.

THE FOUR-STEP IVP METHOD

Transform both sides using the derivative formulas above. Let $Y(s) = \mathcal{L}\{y(t)\}$ .
Substitute the initial conditions immediately — plug in $y(0)$ , $y'(0)$ , and so on right after transforming. Do not carry them as symbols through the algebra.
Solve the resulting algebraic equation for $Y(s)$ . This is now ordinary algebra, not differential equations.
Invert $Y(s)$ back to $y(t)$ using the partial fraction methods from Part 2.

10 Worked Board Exam Problems

Problem 1. Find the Laplace Transform of a First Derivative With Given Initial Condition

Problem: Find $\mathcal{L}\{f'(t)\}$ given that $f(0) = 4$ and $\mathcal{L}\{f(t)\} = \dfrac{3}{s^2+9}$ .

Given: $F(s) = \dfrac{3}{s^2+9}$ , $f(0) = 4$

Find: $\mathcal{L}\{f'(t)\}$

Solution:

Step 1: Apply the first derivative formula directly.

$\mathcal{L}\{f'(t)\} = sF(s) - f(0)$

Step 2: Substitute $F(s)$ and $f(0) = 4$ .

$\mathcal{L}\{f'(t)\} = s\left(\dfrac{3}{s^2+9}\right) - 4$

✓ ANSWER: $\mathcal{L}\{f'(t)\} = \dfrac{3s}{s^2+9} - 4$

Examiner note: This result is already in simplified form — do not force a common denominator unless the problem asks for a single combined fraction. The two-term form here is easier to invert if needed later.

Problem 2. Find the Laplace Transform of a Second Derivative

Problem: Find $\mathcal{L}\{y''(t)\}$ given $y(0) = 2$ , $y'(0) = -1$ , and $Y(s) = \mathcal{L}\{y(t)\}$ .

Given: $y(0) = 2$ , $y'(0) = -1$ , $Y(s)$ is the transform of $y(t)$

Find: $\mathcal{L}\{y''(t)\}$ in terms of $Y(s)$

Solution:

Step 1: Write the second derivative formula.

$\mathcal{L}\{y''(t)\} = s^2Y(s) - sy(0) - y'(0)$

Step 2: Substitute the given initial conditions.

$\mathcal{L}\{y''(t)\} = s^2Y(s) - s(2) - (-1)$

✓ ANSWER: $\mathcal{L}\{y''(t)\} = s^2Y(s) - 2s + 1$

Examiner note: The sign on $y'(0)$ is negative in the formula — $s^2Y(s) - sy(0) - y'(0)$ . When $y'(0) = -1$ , that term becomes $-(-1) = +1$ . Watch the double negative — it is a common arithmetic slip.

Problem 3. Solve a First-Order IVP

Problem: Solve $y' - 3y = 0$ with $y(0) = 2$ .

Given: $y' - 3y = 0$ , $y(0) = 2$

Find: $y(t)$

Solution:

Step 1: Transform both sides. Let $Y(s) = \mathcal{L}\{y(t)\}$ .

$[sY(s) - y(0)] - 3Y(s) = 0$

Step 2: Substitute $y(0) = 2$ .

$sY(s) - 2 - 3Y(s) = 0$

Step 3: Solve algebraically for $Y(s)$ .

$Y(s)(s - 3) = 2 \implies Y(s) = \dfrac{2}{s-3}$

Step 4: Invert using the exponential pair.

✓ ANSWER: $y(t) = 2e^{3t}$

Examiner note: First-order IVPs with zero right-hand side always produce a single exponential term. The initial condition controls the coefficient, the ODE coefficient controls the exponent.

Problem 4. Solve a Second-Order IVP With Real Distinct Roots

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

Given: $y'' - 3y' + 2y = 0$ , $y(0) = 1$ , $y'(0) = 0$

Find: $y(t)$

Solution:

Step 1: Transform both sides, letting $Y(s) = \mathcal{L}\{y(t)\}$ .

$[s^2Y - sy(0) - y'(0)] - 3[sY - y(0)] + 2Y = 0$

Step 2: Substitute initial conditions $y(0) = 1$ , $y'(0) = 0$ .

$[s^2Y - s] - 3[sY - 1] + 2Y = 0$

Step 3: Collect all $Y$ terms and solve.

$(s^2 - 3s + 2)Y = s - 3 \implies Y(s) = \dfrac{s-3}{(s-1)(s-2)}$

Step 4: Decompose using Case 1. Substitute $s = 1$ and $s = 2$ .

$s-3 = A(s-2) + B(s-1)$

$s=1: -2 = -A \implies A = 2$

$s=2: -1 = B \implies B = -1$

Step 5: Invert.

$Y(s) = \dfrac{2}{s-1} - \dfrac{1}{s-2}$

✓ ANSWER: $y(t) = 2e^{t} - e^{2t}$

Examiner note: The four-step pattern applies here exactly as described — transform, substitute, solve for $Y(s)$ , invert. Steps 4 and 5 are just Part 2 work applied to the result of Step 3.

Problem 5. Solve a Second-Order IVP With a Non-Zero Right-Hand Side

Problem: Solve $y'' + 4y = 8$ with $y(0) = 0$ , $y'(0) = 0$ .

Given: $y'' + 4y = 8$ , $y(0) = 0$ , $y'(0) = 0$

Find: $y(t)$

Solution:

Step 1: Transform both sides. The right-hand side $8$ transforms to $8/s$ .

$[s^2Y - 0 - 0] + 4Y = \dfrac{8}{s}$

Step 2: With zero initial conditions, the equation simplifies cleanly.

$(s^2 + 4)Y = \dfrac{8}{s} \implies Y(s) = \dfrac{8}{s(s^2+4)}$

Step 3: Decompose — linear factor $s$ gets constant $A$ , quadratic factor gets $Bs+C$ .

$\dfrac{8}{s(s^2+4)} = \dfrac{A}{s} + \dfrac{Bs+C}{s^2+4}$

$8 = A(s^2+4) + (Bs+C)s$

$s=0: 8 = 4A \implies A = 2$

$s^2: 0 = A + B \implies B = -2$

$s^1: 0 = C$

Step 4: Invert.

$Y(s) = \dfrac{2}{s} - \dfrac{2s}{s^2+4}$

✓ ANSWER: $y(t) = 2 - 2\cos(2t)$

Examiner note: When initial conditions are all zero, the transform of the ODE simplifies dramatically. Every $-sy(0)$ and $-y'(0)$ term drops to zero, and you go straight from the transformed equation to solving for $Y(s)$ .

Problem 6. Solve an IVP With a Complex Root Denominator

Problem: Solve $y'' + 2y' + 5y = 0$ with $y(0) = 1$ , $y'(0) = 3$ .

Given: $y'' + 2y' + 5y = 0$ , $y(0) = 1$ , $y'(0) = 3$

Find: $y(t)$

Solution:

Step 1: Transform and substitute initial conditions.

$[s^2Y - s - 3] + 2[sY - 1] + 5Y = 0$

$(s^2 + 2s + 5)Y = s + 3 + 2 = s + 5$

Step 2: Solve for $Y(s)$ .

$Y(s) = \dfrac{s+5}{s^2+2s+5}$

Step 3: Complete the square in the denominator to match the shifted pair form.

$s^2 + 2s + 5 = (s+1)^2 + 4$

Step 4: Rewrite the numerator to split into shifted cosine and sine pieces.

$Y(s) = \dfrac{s+5}{(s+1)^2+4} = \dfrac{(s+1) + 4}{(s+1)^2+4}$

$= \dfrac{s+1}{(s+1)^2+4} + \dfrac{4}{(s+1)^2+4} = \dfrac{s+1}{(s+1)^2+4} + 2\cdot\dfrac{2}{(s+1)^2+4}$

Step 5: Invert using the first shifting theorem in reverse — these are shifted cosine and sine pairs with $a = -1$ , $k = 2$ .

✓ ANSWER: $y(t) = e^{-t}\cos(2t) + 2e^{-t}\sin(2t)$

Examiner note: Completing the square is the tool for complex roots — it forces the denominator into the $(s-a)^2 + k^2$ form that matches the shifted sine and cosine pairs. If the denominator discriminant is negative, completing the square is always the next step.

Problem 7. Use the Integration Property

Problem: Given $\mathcal{L}\{f(t)\} = \dfrac{1}{s+2}$ , find $\mathcal{L}\left\{\int_0^t f(\tau)\,d\tau\right\}$ .

Given: $F(s) = \dfrac{1}{s+2}$

Find: $\mathcal{L}\left\{\int_0^t f(\tau)\,d\tau\right\}$

Solution:

Step 1: Apply the integration property directly.

$\mathcal{L}\left\{\int_0^t f(\tau)\,d\tau\right\} = \dfrac{F(s)}{s}$

Step 2: Substitute $F(s)$ .

$= \dfrac{1}{s(s+2)}$

✓ ANSWER: $\dfrac{1}{s(s+2)}$

Examiner note: This is one of the faster problem types in the series — the integration property is a single division by $s$ , no other work required. The result here could be inverted further using Part 2’s Case 1 decomposition if the problem asked for the time-domain answer.

Problem 8. Apply the Second Shifting Theorem

Problem: Find the Laplace Transform of $f(t) = (t-2)^2 u(t-2)$ .

Given: $f(t) = (t-2)^2 u(t-2)$

Find: $\mathcal{L}\{f(t)\}$

Solution:

Step 1: Recognize the second shifting theorem pattern $f(t-a)u(t-a)$ with $a = 2$ and $g(t) = t^2$ .

$\mathcal{L}\{g(t-2)u(t-2)\} = e^{-2s}G(s)$

Step 2: Find $G(s) = \mathcal{L}\{t^2\}$ using the basic pair.

$G(s) = \dfrac{2!}{s^3} = \dfrac{2}{s^3}$

Step 3: Apply the shift.

✓ ANSWER: $\mathcal{L}\{(t-2)^2 u(t-2)\} = \dfrac{2e^{-2s}}{s^3}$

Examiner note: The function inside the unit step must be written in the form $g(t-a)$ — not $g(t)$ — before applying the theorem. If the problem gives you $t^2 u(t-2)$ instead of $(t-2)^2 u(t-2)$ , you need to rewrite $t^2$ in terms of $(t-2)$ first by expanding $((t-2)+2)^2$ .

Problem 9. Solve an IVP With a Forcing Function on the Right Side

Problem: Solve $y'' + y = \sin(t)$ with $y(0) = 0$ , $y'(0) = 0$ .

Given: $y'' + y = \sin(t)$ , $y(0) = 0$ , $y'(0) = 0$

Find: $y(t)$

Solution:

Step 1: Transform both sides. The forcing function $\sin(t)$ transforms to $1/(s^2+1)$ .

$s^2Y + Y = \dfrac{1}{s^2+1}$

Step 2: Factor and solve for $Y(s)$ .

$(s^2+1)Y = \dfrac{1}{s^2+1} \implies Y(s) = \dfrac{1}{(s^2+1)^2}$

Step 3: Recognize this as the standard repeated quadratic pair from Part 2, Problem 8.

$\mathcal{L}^{-1}\left\{\dfrac{1}{(s^2+1)^2}\right\} = \dfrac{1}{2}(\sin t - t\cos t)$

✓ ANSWER: $y(t) = \dfrac{1}{2}(\sin t - t\cos t)$

Examiner note: This is the resonance case — the forcing frequency matches the natural frequency of the system, producing the $t\cos(t)$ term that grows without bound. On EE and ECE boards, this pattern shows up in RLC circuit resonance problems, not just as an abstract ODE.

Problem 10. Solve a Full IVP With a Step-Function Input (ME/EE Application)

Problem: A mechanical system is governed by $x'' + 5x' + 6x = u(t)$ with $x(0) = 0$ , $x'(0) = 0$ , where $u(t)$ is the unit step function. Find $x(t)$ .

Given: $x'' + 5x' + 6x = u(t)$ , $x(0) = 0$ , $x'(0) = 0$

Find: $x(t)$

Solution:

Step 1: Transform both sides. The unit step $u(t)$ transforms to $1/s$ . Initial conditions are zero so those terms drop out.

$(s^2 + 5s + 6)X(s) = \dfrac{1}{s}$

Step 2: Solve for $X(s)$ and factor the denominator.

$X(s) = \dfrac{1}{s(s^2+5s+6)} = \dfrac{1}{s(s+2)(s+3)}$

Step 3: Decompose using Case 1 — three distinct linear factors.

$\dfrac{1}{s(s+2)(s+3)} = \dfrac{A}{s} + \dfrac{B}{s+2} + \dfrac{C}{s+3}$

$s=0: A = \dfrac{1}{6}, \quad s=-2: B = \dfrac{1}{(-2)(1)} = -\dfrac{1}{2}, \quad s=-3: C = \dfrac{1}{(-3)(-1)} = \dfrac{1}{3}$

Step 4: Invert each term.

✓ ANSWER: $x(t) = \dfrac{1}{6} - \dfrac{1}{2}e^{-2t} + \dfrac{1}{3}e^{-3t}$

Examiner note: The unit step input $u(t)$ on the right-hand side models a sudden constant force or voltage applied at $t = 0$ . The $1/s$ term it produces in the transform creates the constant term $1/6$ in the final answer, representing the steady-state response of the system.

Common Mistakes and Examiner Traps

❌ Common Mistake	✅ Correct Approach
Carrying initial condition symbols through the algebra instead of substituting immediately	Substitute $y(0)$ , $y'(0)$ right after transforming, before doing any algebra on the $Y(s)$ terms. Symbols in the algebra introduce sign errors.
Getting the sign wrong on $y'(0)$ when it is negative — missing the double negative in $-y'(0)$	The second derivative formula is $s^2Y - sy(0) - y'(0)$ . When $y'(0) = -1$ , the term becomes $-(-1) = +1$ . Write it out explicitly rather than doing it mentally.
Forgetting to transform the right-hand side of the ODE	Transform both sides. A constant $c$ on the right becomes $c/s$ , $\sin(kt)$ becomes $k/(s^2+k^2)$ , and so on using the Part 1 table.
Applying the second shifting theorem to $g(t)u(t-a)$ instead of $g(t-a)u(t-a)$	The function inside the step must be shifted by $a$ also. Rewrite $g(t)$ as a function of $(t-a)$ first, then apply the theorem. The two forms are not equivalent.
Stopping after finding $Y(s)$ without inverting back to $y(t)$	The IVP asks for $y(t)$ , not $Y(s)$ . Always complete Step 4. $Y(s)$ is an intermediate result, not the final answer.
Skipping the completing-the-square step when the denominator has complex roots	If $s^2 + bs + c$ has a negative discriminant, complete the square to put it in $(s+p)^2 + q^2$ form before inverting. This is the only way to match the shifted cosine and sine pairs.

Board Exam Quick Tips

Substitute initial conditions immediately after transforming, not at the end. Carrying $y(0)$ and $y'(0)$ as symbols through the algebra is the single fastest way to introduce a sign error on this topic.
The number of initial conditions equals the order of the ODE. A second-order ODE needs exactly two initial conditions — $y(0)$ and $y'(0)$ . No more, no less.
After solving for $Y(s)$ , you are back in Part 2 territory. Apply the same decomposition cases — the ODE context does not change the partial fraction method.
Zero initial conditions simplify the transform dramatically. Every initial condition term drops out, and you go straight from the ODE to $(characteristic\ polynomial) \cdot Y(s) = \mathcal{L}\{RHS\}$ .
If the denominator of $Y(s)$ has a negative discriminant, complete the square. This forces the denominator into shifted sine/cosine form and is the only clean path to the inverse in that case.

Frequently Asked Questions

Q1. Why is the Laplace method faster than classical methods for ODEs?

Classical methods require solving a homogeneous equation, finding a particular solution for the forcing term, and then combining both with the initial conditions. The Laplace method folds all of this into one algebraic equation for $Y(s)$ — initial conditions get plugged in during the transform step, the forcing function appears automatically on the right side, and the inversion at the end gives the complete solution directly.

Q2. What if the ODE has a variable coefficient like $t \cdot y'$ ?

Constant coefficient ODEs are the type that transform cleanly into algebraic equations for $Y(s)$ . Variable coefficient ODEs like $t \cdot y' + y = 0$ require the multiplication-by- $t$ property of the Laplace Transform, which introduces $-dF/ds$ terms. These are less common on the boards but can appear as advanced items on EE and ECE exams.

Q3. What is the difference between the first and second shifting theorems?

The first shifting theorem from Part 1 handles multiplication by $e^{at}$ in the time domain, which shifts $F(s)$ along the $s$ -axis. The second shifting theorem in this post handles time-domain translation — shifting a function to start at $t = a$ instead of $t = 0$ — using the unit step function and an $e^{-as}$ factor in the $s$ -domain.

Q4. Can I use the Laplace method on a system of ODEs?

Yes, and this appears on EE and ECE boards in coupled circuit problems. Transform each equation in the system, substitute initial conditions, and solve the resulting system of algebraic equations for $X_1(s)$ , $X_2(s)$ , and so on using standard elimination or substitution. Each $X_i(s)$ is then inverted separately.

Q5. When does completing the square produce complex exponentials versus real ones?

If after completing the square the denominator is $(s+p)^2 + q^2$ with $q^2 > 0$ , the inverse involves $e^{-pt}\sin(qt)$ and $e^{-pt}\cos(qt)$ — oscillatory terms. If the denominator factors into real linear terms $(s+p)(s+r)$ , the inverse involves real exponentials $e^{-pt}$ and $e^{-rt}$ . The sign of the discriminant of the original quadratic tells you which case you are in before you start.

What is Next

You now have the full IVP toolkit — transform an ODE, substitute initial conditions, solve for $Y(s)$ , invert to $y(t)$ . In Part 4, the content shifts to three special topics that require their own framework: the Dirac delta function for impulse inputs, periodic waveform transforms, and the convolution theorem. These three tools are the specific additions that make the EE and ECE board’s circuit and signals problems solvable without evaluating difficult integrals directly.

→ Continue to Part 4 — Impulse Functions, Periodic Functions, and Convolution

→ Back to the Laplace Transform Series Index

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Laplace Transform Derivatives, Integrals & Solving ODEs | Board Exam Reviewer

Transform of Derivatives

Transform of an Integral

The Second Shifting Theorem — Unit Step Functions

The Four-Step IVP Pattern

10 Worked Board Exam Problems

Problem 1. Find the Laplace Transform of a First Derivative With Given Initial Condition

Problem 2. Find the Laplace Transform of a Second Derivative

Problem 3. Solve a First-Order IVP

Problem 4. Solve a Second-Order IVP With Real Distinct Roots

Problem 5. Solve a Second-Order IVP With a Non-Zero Right-Hand Side

Problem 6. Solve an IVP With a Complex Root Denominator

Problem 7. Use the Integration Property

Problem 8. Apply the Second Shifting Theorem

Problem 9. Solve an IVP With a Forcing Function on the Right Side

Problem 10. Solve a Full IVP With a Step-Function Input (ME/EE Application)

Common Mistakes and Examiner Traps

Board Exam Quick Tips

Frequently Asked Questions

What is Next

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Transform of Derivatives

Transform of an Integral

The Second Shifting Theorem — Unit Step Functions

The Four-Step IVP Pattern

10 Worked Board Exam Problems

Problem 1. Find the Laplace Transform of a First Derivative With Given Initial Condition

Problem 2. Find the Laplace Transform of a Second Derivative

Problem 3. Solve a First-Order IVP

Problem 4. Solve a Second-Order IVP With Real Distinct Roots

Problem 5. Solve a Second-Order IVP With a Non-Zero Right-Hand Side

Problem 6. Solve an IVP With a Complex Root Denominator

Problem 7. Use the Integration Property

Problem 8. Apply the Second Shifting Theorem

Problem 9. Solve an IVP With a Forcing Function on the Right Side

Problem 10. Solve a Full IVP With a Step-Function Input (ME/EE Application)

Common Mistakes and Examiner Traps

Board Exam Quick Tips

Frequently Asked Questions

What is Next

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