# Laplace Transforms

Physics I · Differential Equations and Vector Calculus · https://tryals.app/learn/physics-i/laplace-transforms

The **Laplace transform** converts a hard differential equation into a simple algebraic equation by mapping functions of time into the frequency domain.

**Definition**

$$\mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st}f(t)\,dt$$

Where $t$ is time, $s$ is the complex transform variable, $f(t)$ is the input function, and $F(s)$ is its transform.

**Essential Transform Pairs**

| $f(t)$ | $F(s)$ |
| --- | --- |
| $1$ | $1/s$ |
| $t^n$ | $n!/s^{n+1}$ |
| $e^{at}$ | $1/(s-a)$ |
| $\sin(\omega t)$ | $\omega/(s^{2}+\omega^{2})$ |
| $\cos(\omega t)$ | $s/(s^{2}+\omega^{2})$ |

**Key Properties**

- **Linearity:** $\mathcal{L}\{af+bg\} = aF + bG$
- **First Derivative:** $\mathcal{L}\{f'\} = sF(s) - f(0)$
- **Second Derivative:** $\mathcal{L}\{f''\} = s^{2}F(s) - sf(0) - f'(0)$

## Solving with Laplace

**Worked Example**: Solve $y'' + 4y = 0$ with $y(0)=1$ and $y'(0)=0$.

1. Transform both sides: $(s^{2}Y - s - 0) + 4Y = 0$.
2. Solve algebraically for $Y$: $Y = \dfrac{s}{s^{2}+4}$.
3. Take the inverse transform: $y = \cos(2t)$.

**Partial fractions** are required to invert complex $F(s)$ expressions back into time-domain functions $f(t)$.

**Physics Link**: Essential for analyzing **linear circuits** and **control systems**, where initial conditions feed directly into the transformed equation.

**Common Pitfall**: Laplace transforms absorb initial conditions immediately via $\mathcal{L}\{y'\} = sY - y(0)$. Bolting them on at the end like standard ODEs will double-count or drop your values.

## Practice questions

5 of this lesson's 11 practice questions, with answers. The full set is in the app.

### 1. Which properties does the Laplace transform have?

A. L{f'(t)} = sF(s) − $f(0)$
B. Linearity: L{af+bg} = aL{f}+bL{g}
C. $\mathcal{L}\{f\cdot g\} = F \cdot G$ (product becomes product)
D. L{$e^{at} f(t)$} = $F(s-a)$ (frequency shift)

**Answer:** A. L{f'(t)} = sF(s) − $f(0)$; B. Linearity: L{af+bg} = aL{f}+bL{g}; D. L{$e^{at} f(t)$} = $F(s-a)$ (frequency shift)

**Why:** The transform of a product is a convolution, not a simple product.

Page: https://tryals.app/practice/physics-i/laplace-transforms/which-properties-does-the-laplace-transform-have

### 2. Sort these into t-domain or s-domain operations.

**Answer:**

- t-domain (time): Differentiation d/dt, Convolution f * g, Initial value $f(0)$
- s-domain (frequency): Multiplication by s, Product $F(s)$·$G(s)$, Partial fraction decomposition

**Why:** The Laplace transform maps between time and frequency domains.

Page: https://tryals.app/practice/physics-i/laplace-transforms/sort-these-into-t-domain-or-s-domain-operations

### 3. Order the steps of solving an initial-value problem with the **Laplace transform**.

**Answer:**

1. Transform both sides, turning derivatives into multiplications by $s$
2. Insert the initial conditions the transform absorbed
3. Solve the resulting **algebraic** equation for $Y(s)$
4. Split $Y(s)$ into partial fractions
5. Invert each simple piece back to the time domain

**Why:** Laplace converts an ODE into algebra, bakes the initial conditions in from the start, and a table lookup brings the answer home. Its real superpower: discontinuous inputs (switches, impulses) that defeat classical methods are handled effortlessly.

Page: https://tryals.app/practice/physics-i/laplace-transforms/order-the-steps-of-solving-an-initial-value-problem-with-the-laplace

### 4. The Laplace transform converts a differential equation into an algebraic one. What does that exchange cost?

A. Nothing: the method is strictly easier at every step
B. The solution arrives transformed and must be inverted
C. Accuracy, since the transform is an approximation
D. Generality, since the method works only for linear equations with constant coefficients

**Answer:** B. The solution arrives transformed and must be inverted

**Why:** The work relocates rather than vanishing. Partial fractions and transform tables exist to make the return trip manageable, and that step is where the effort now sits.

Page: https://tryals.app/practice/physics-i/laplace-transforms/the-laplace-transform-converts-a-differential-equation-into-an

### 5. The Laplace transform converts a differential equation into an algebraic equation.

**Answer:** True

**Why:** L{f'(t)} = sF(s) − $f(0)$, turning derivatives into polynomial operations.

Page: https://tryals.app/practice/physics-i/laplace-transforms/the-laplace-transform-converts-a-differential-equation-into-an-2