The Laplace transform is a clever tool that converts a differential equation into an ordinary algebra problem. You transform the equation, solve for the unknown using algebra, then transform back to get the answer, with the initial conditions built in automatically. Engineers use it constantly for circuits and control systems, and it is a favourite topic in university exams.
The Definition
For a function \( f(t) \) defined for \( t \geq 0 \), the Laplace transform is
$$ \mathcal{L}\{f(t)\} = F(s) = \int_0^\infty e^{-st}f(t)\,dt, $$
wherever the integral converges. The transform takes a function of time \( t \) and produces a function of a new variable \( s \).
Worked Example 1: Deriving a Basic Transform
Find \( \mathcal{L}\{e^{at}\} \).
Step 1: Write the integral.
$$ \mathcal{L}\{e^{at}\} = \int_0^\infty e^{-st}e^{at}\,dt = \int_0^\infty e^{-(s-a)t}\,dt. $$
Step 2: Integrate. For \( s > a \),
$$ \int_0^\infty e^{-(s-a)t}\,dt = \left[\frac{-e^{-(s-a)t}}{s-a}\right]_0^\infty = 0 - \left(\frac{-1}{s-a}\right) = \frac{1}{s-a}. $$
So \( \mathcal{L}\{e^{at}\} = \frac{1}{s-a} \) for \( s > a \). Setting \( a = 0 \) gives \( \mathcal{L}\{1\} = \frac{1}{s} \).
The Essential Table
You rarely compute integrals in exams. Instead, you use a table like this:
| \( f(t) \) | \( F(s) \) |
|---|---|
| \( 1 \) | \( \frac{1}{s} \) |
| \( t^n \) | \( \frac{n!}{s^{n+1}} \) |
| \( e^{at} \) | \( \frac{1}{s-a} \) |
| \( \sin bt \) | \( \frac{b}{s^2+b^2} \) |
| \( \cos bt \) | \( \frac{s}{s^2+b^2} \) |
Key Properties
- Linearity: \( \mathcal{L}\{af + bg\} = aF(s) + bG(s) \).
- First shifting theorem: \( \mathcal{L}\{e^{at}f(t)\} = F(s - a) \). For example, \( \mathcal{L}\{e^{2t}\sin 3t\} = \frac{3}{(s-2)^2 + 9} \).
- Transform of derivatives (the property that makes everything work):
$$ \mathcal{L}\{y'\} = sY(s) - y(0), $$
$$ \mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0). $$
These formulas turn derivatives into multiplication by \( s \), with the initial values included.
The Method for Solving IVPs
- Take the Laplace transform of both sides of the ODE.
- Substitute the initial conditions using the derivative formulas.
- Solve algebraically for \( Y(s) \).
- Use partial fractions to split \( Y(s) \) into table forms.
- Take the inverse transform to find \( y(t) \).
Worked Example 2: A First-Order IVP
Solve \( y' + 2y = e^{-t} \) with \( y(0) = 3 \).
Step 1: Transform both sides.
$$ \big(sY - y(0)\big) + 2Y = \frac{1}{s+1}. $$
Step 2: Substitute \( y(0) = 3 \).
$$ sY - 3 + 2Y = \frac{1}{s+1}. $$
Step 3: Solve for \( Y \).
$$ (s+2)Y = 3 + \frac{1}{s+1} \quad\Longrightarrow\quad Y = \frac{3}{s+2} + \frac{1}{(s+1)(s+2)}. $$
Step 4: Partial fractions. Write \( \frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2} \). Then \( 1 = A(s+2) + B(s+1) \). Putting \( s = -1 \) gives \( A = 1 \); putting \( s = -2 \) gives \( B = -1 \). So
$$ Y = \frac{3}{s+2} + \frac{1}{s+1} - \frac{1}{s+2} = \frac{1}{s+1} + \frac{2}{s+2}. $$
Step 5: Invert.
$$ y(t) = e^{-t} + 2e^{-2t}. $$
Check: \( y(0) = 1 + 2 = 3 \). Also \( y' = -e^{-t} - 4e^{-2t} \), so \( y' + 2y = -e^{-t} - 4e^{-2t} + 2e^{-t} + 4e^{-2t} = e^{-t} \). Correct.
Worked Example 3: A Second-Order IVP
Solve \( y'' + 4y = 8 \) with \( y(0) = 0 \) and \( y'(0) = 0 \).
Step 1: Transform.
$$ s^2Y - sy(0) - y'(0) + 4Y = \frac{8}{s}. $$
Step 2: Substitute the zero initial conditions.
$$ s^2Y + 4Y = \frac{8}{s}. $$
Step 3: Solve for \( Y \).
$$ Y = \frac{8}{s\left(s^2+4\right)}. $$
Step 4: Partial fractions. Write
$$ \frac{8}{s(s^2+4)} = \frac{A}{s} + \frac{Bs + C}{s^2+4}. $$
Multiply out: \( 8 = A(s^2+4) + (Bs + C)s \). Putting \( s = 0 \) gives \( A = 2 \). Comparing the \( s^2 \) terms, \( 0 = A + B \), so \( B = -2 \). Comparing the \( s \) terms, \( C = 0 \). Hence
$$ Y = \frac{2}{s} - \frac{2s}{s^2+4}. $$
Step 5: Invert using the table.
$$ y(t) = 2 - 2\cos 2t. $$
Check: \( y'' = 8\cos 2t \), so \( y'' + 4y = 8\cos 2t + 8 - 8\cos 2t = 8 \). Also \( y(0) = 0 \) and \( y'(0) = 4\sin 0 = 0 \). Correct.
Common Mistakes
- Dropping initial conditions in the derivative formulas, especially the \( -sy(0) \) term for \( y'' \).
- Sign errors with shifts: \( \mathcal{L}\{e^{-3t}\} = \frac{1}{s+3} \), not \( \frac{1}{s-3} \).
- Mixing up sine and cosine: \( \frac{s}{s^2+b^2} \) is cosine; \( \frac{b}{s^2+b^2} \) is sine.
- Weak partial fractions. Most of the work, and most of the errors, happen here, so check by recombining.
Key Takeaways
- The Laplace transform \( F(s) = \int_0^\infty e^{-st}f(t)\,dt \) turns ODEs into algebra.
- Learn the basic table and the derivative formulas by heart.
- Initial conditions enter automatically, so there are no constants to find at the end.
- Partial fractions are the bridge back to table forms.
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