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Differential Equations Unit 7: Laplace transforms

The Laplace transform turns a differential equation into an algebra problem. Solve for the transform, then invert.

Covers: Definition, inverse transforms, solving IVPs, step functions

Study guide

The Laplace transform turns a differential equation into an algebra problem. It's defined by L{f(t)} = F(s) = ∫₀^∞ e^(−st) f(t) dt. Standard transforms include L{1} = 1/s, L{tⁿ} = n!/sⁿ⁺¹, L{eᵃᵗ} = 1/(s − a), L{sin bt} = b/(s² + b²), and L{cos bt} = s/(s² + b²). The transform is linear, so constants and sums pass through.

The key property handles derivatives: L{y′} = sY(s) − y(0) and L{y″} = s²Y(s) − sy(0) − y′(0). Initial conditions are built in automatically. The first shifting theorem says L{eᵃᵗf(t)} = F(s − a).

To solve an initial value problem, transform both sides of the equation, plug in the initial conditions, solve algebraically for Y(s), and then find y(t) with the inverse transform. Inverse transforms usually require partial fractions or completing the square so each piece matches a table entry.

Step functions model forces that switch on or off. The unit step u_c(t) is 0 before t = c and 1 after. The second shifting theorem says L{u_c(t)f(t − c)} = e^(−cs)F(s), so a factor e^(−cs) in a transform means a delay: the function turns on at t = c. The Dirac delta δ(t − c) models an instantaneous impulse, with L{δ(t − c)} = e^(−cs). Convolution gives a product rule for transforms: L{f * g} = F(s)G(s).

For p(D)y = f(t) with zero initial conditions, Y(s) = W(s)F(s), where W(s) = 1/p(s) is the transfer function. The weight function (impulse response) w(t) = L⁻¹{W(s)} is the response to δ(t), and every other response is the convolution y = w * f. The convolution of f and g is (f * g)(t) = ∫₀ᵗ f(τ)g(t − τ) dτ. It is commutative, and its transform is F(s)G(s), so it can be computed either directly or by inverting the product.

In the generalized sense, the derivative of u_c(t) is δ(t − c). So a function with a jump of size J at t = c has a derivative with an impulse J·δ(t − c) at that point.

Key ideas

• L{f(t)} = ∫₀^∞ e^(−st) f(t) dt.
• Derivatives become multiplication by s, with initial conditions built in.
• Solve for Y(s), then use partial fractions and a table to invert.
• Step functions model switches turning on; shifting theorems handle them.

Formulas and rules

Common mistakes

• Dropping initial conditions in L{y″}.
• Inverting before doing partial fractions.

Words to know

Definition of the Laplace transform

L{f} = ∫₀^∞ e^(−st) f(t) dt

L{eᵃᵗ}

1/(s − a)

L{y′}

sY(s) − y(0)

L{sin bt}

b/(s² + b²)

e^(−cs) in a Laplace transform

A time shift: the function turns on at t = c.

Laplace transform

L{f} = ∫₀^∞ e^(−st) f(t) dt

L{1}

1/s

L{tⁿ}

n!/sⁿ⁺¹

L{cos bt}

s/(s² + b²)

L{y″}

s²Y(s) − sy(0) − y′(0)

Unit step function u_c(t)

0 before t = c, 1 after.

Dirac delta

An instantaneous impulse; L{δ(t − c)} = e^(−cs).

Convolution theorem

L{f * g} = F(s)G(s)

Solving an IVP with Laplace transforms

Transform, plug in initial conditions, solve for Y(s), invert with partial fractions.

Sample practice questions

21 more questions like these, from level 1 to 5, are in the practice set.