Guides / Differential equations

Laplace transforms: transform the equation with its initial values, solve for Y(s), then invert piece by piece from the table

The Laplace transform F(s) = ∫₀^∞ e^−st f(t) dt turns a function of time into a function of s, and its point is that it turns a differential equation into algebra. A derivative becomes multiplication by s, minus the starting value, so an initial-value problem becomes one linear equation for Y(s) with the initial conditions already inside it. Solve for Y, split it into partial fractions that each appear in a short table, and read y(t) back off the table. Most wrong answers come from dropping the initial values, shifting the wrong thing, or using the wrong partial-fraction form.

Why it works

Integrate e^−sty′(t) by parts: the boundary term at 0 gives −y(0), the one at infinity vanishes for large enough s, and what is left is s times the transform of y. So L{y′} = sY − y(0), and applying it twice gives the rule for y″. Differentiation has become multiplication, and the starting values ride along as constants.

The transform is linear, so a sum of table entries comes back as the same sum of functions, and two different continuous functions never share a transform, so reading the answer off the table is safe. Partial fractions are the step that breaks a rational Y(s) into those table entries. A factor e^at in time slides the whole transform to s − a, and a delay by c in time multiplies the transform by e^−cs.

Example 1: a transform from the table

Find the transform of f(t) = 5 + 2t² − 3e^−4t + e^t sin 2t.

1. L{5} = 5/sFive times the transform of 1.
2. L{2t²} = 2 · 2!/s³ = 4/s³The power rule with n = 2. The factorial is easy to forget.
3. L{−3e^−4t} = −3/(s + 4)e^at with a = −4.
4. L{sin 2t} = 2/(s² + 4)b on top, b² below.
5. L{e^t sin 2t}The factor e^t means a = 1: replace every s by s − 1.
6. = 2/((s − 1)² + 4)The shifted sine.
7. F = 5/s + 4/s³Add the pieces: F(s) = 5/s + 4/s³ − 3/(s + 4) + 2/((s − 1)² + 4), valid for s > 1, where every piece converges.

Example 2: an inverse with a repeated factor

Find the function whose transform is F(s) = (s + 3)/((s + 1)²(s + 2)).

1. A/(s + 1) + B/(s + 1)²A squared factor needs both powers, plus C/(s + 2) for the other factor.
2. s + 3 = A(s + 1)(s + 2)Multiply through by the denominator; the right side continues + B(s + 2) + C(s + 1)².
3. s = −1: 2 = BThe A and C terms vanish.
4. s = −2: 1 = CThe A and B terms vanish.
5. A + C = 0, so A = −1Compare the s² terms: the left side has none.
6. 1/(s + 1)² → te^−t1/s² → t, shifted by a = −1.
7. f = −e^−t + 2te^−t + e^−2tCheck: f(0) = −1 + 0 + 1 = 0, matching sF(s) → 0 as s → ∞.

Example 3: a second-order initial-value problem

Solve y″ − y′ − 6y = 0 with y(0) = 1 and y′(0) = 8.

1. s²Y − s − 8The transform of y″: s²Y − s y(0) − y′(0).
2. − (sY − 1) − 6Y = 0The transforms of −y′ and −6y, carried on the same line of the equation.
3. (s² − s − 6)Y = s + 7Collect Y; the constants s + 8 − 1 move to the right.
4. Y = (s + 7)/((s − 3)(s + 2))Factor the quadratic and divide.
5. = 2/(s − 3) − 1/(s + 2)Cover-up: at s = 3, 10/5 = 2; at s = −2, 5/(−5) = −1.
6. y = 2e^3t − e^−2tCheck: y(0) = 2 − 1 = 1 and y′(0) = 6 + 2 = 8. The characteristic roots 3 and −2 appear as the denominators’ roots, as they must.

Example 4: forcing that switches on later

Solve y′ + y = u(t − 1) with y(0) = 0: a constant input of 1 switches on at t = 1. Find y(2).

1. sY + Y = e^−s/sL{u(t − 1)} = e^−s/s, and y(0) = 0.
2. Y = e^−s · 1/(s(s + 1))Set the exponential aside; it only delays the answer.
3. 1/(s(s + 1)) = 1/s − 1/(s + 1)Cover-up: 1 at s = 0, −1 at s = −1.
4. 1/s − 1/(s + 1) → 1 − e^−tThe undelayed response.
5. y = u(t − 1)(1 − e^{−(t − 1)})The second shift theorem: replace t by t − 1 everywhere, and switch on at t = 1.
6. y(2) = 1 − e⁻¹ ≈ 0.632The solution is 0 up to t = 1, then rises toward 1 with time constant 1.

Common mistakes, and the exact line they happen on

1. Dropping the initial values from the derivative rule

The equation of Example 3.

1. y″ − y′ − 6y = 0Right so far.
2. (s² − s − 6)Y = 0This is the mistake. This used y″ → s²Y and y′ → sY, which hold only when y(0) = y′(0) = 0. It forces Y = 0 and y = 0, which fails y(0) = 1. In a homogeneous equation the initial values are the only thing driving the solution: the right side should be s + 7.

2. Shifting the outside but not the inside

The delayed input of Example 4.

1. 1/s − 1/(s + 1) → 1 − e^−tRight so far.
2. y = u(t − 1)(1 − e^−t)This is the mistake. The factor e^−s delays the whole function, so every t inside becomes t − 1. This version jumps to 1 − e⁻¹ ≈ 0.632 at the instant t = 1, which a first-order system starting at 0 cannot do, and gives y(2) ≈ 0.865 instead of 0.632.

3. Leaving out a power of a repeated factor

The inverse of Example 2.

1. (s + 3)/((s + 1)²(s + 2))Right so far.
2. B/(s + 1)² + C/(s + 2)This is the mistake. A repeated factor needs every power up to its multiplicity, so the A/(s + 1) term is missing. Cover-up still gives B = 2 and C = 1, but recombining gives 2(s + 2) + (s + 1)² = s² + 4s + 5, not s + 3. The resulting 2te^−t + e^−2t starts at 1 instead of 0.

Practice

1. Find L{e^−3t}.

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   1/(s + 3), for s > −3.

2. Find L{cos 5t + t}.

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   s/(s² + 25) + 1/s².

3. Invert 2/(s + 1)³.

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   t²e^−t: 2/s³ is the transform of t², shifted by a = −1.

4. Invert (s + 1)/(s² + 2s + 10).

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   The denominator is (s + 1)² + 9, so the answer is e^−t cos 3t.

5. Solve y′ − 2y = 0, y(0) = 3, with the Laplace transform.

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   sY − 3 − 2Y = 0, so Y = 3/(s − 2) and y = 3e^2t.