11.3 Solving Differential Equations with Laplace Transforms

Laplace Transforms for Initial Value Problems

Laplace transforms convert differential equations into algebraic equations, making them far easier to solve. Instead of working with derivatives directly, you transform the problem into the s-domain, solve with algebra, then transform back. This approach is especially powerful because it bakes initial conditions right into the process and handles discontinuous forcing functions that would be painful to deal with using classical methods.

Fundamentals of Laplace Transforms

The Laplace transform converts a function of time $f(t)$ into a function of the complex frequency variable $s$:

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

Two properties make this useful for differential equations:

• Linearity: $\mathcal{L}\{af(t) + bg(t)\} = aF(s) + bG(s)$, where $a$ and $b$ are constants
• Derivative transforms: These are what actually let you convert a DE into algebra. The key formulas are:
  • $\mathcal{L}\{f'(t)\} = sF(s) - f(0)$
  • $\mathcal{L}\{f''(t)\} = s^2F(s) - sf(0) - f'(0)$

Notice how the initial conditions $f(0)$ and $f'(0)$ appear directly in the derivative formulas. That's why you don't need to solve for them separately at the end.

Application to Initial Value Problems

The general strategy converts an IVP into an algebraic equation in $s$, solves for $Y(s)$, then inverts back to get $y(t)$. Here's the step-by-step process:

1. Take the Laplace transform of both sides of the differential equation, substituting the derivative formulas and plugging in your initial conditions.
2. Solve for $Y(s)$ algebraically (collect terms, factor, isolate $Y(s)$).
3. Decompose $Y(s)$ using partial fractions if needed, so each piece matches a known inverse transform.
4. Apply the inverse Laplace transform to each term to recover $y(t)$.

This works for constant, exponential, sinusoidal, and many other types of forcing functions.

Worked Example

Step 1: Take the Laplace transform of both sides. Using $\mathcal{L}\{y''\} = s^2Y(s) - sy(0) - y'(0)$ and $\mathcal{L}\{\sin(3t)\} = \frac{3}{s^2+9}$:

$s^2Y(s) - s(1) - 0 + 4Y(s) = \frac{6}{s^2 + 9}$

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

$(s^2 + 4)Y(s) = s + \frac{6}{s^2 + 9}$

$Y(s) = \frac{s}{s^2 + 4} + \frac{6}{(s^2 + 4)(s^2 + 9)}$

Step 3: Use partial fractions on the second term. Since $s^2 + 9$ and $s^2 + 4$ differ by 5:

$\frac{6}{(s^2+4)(s^2+9)} = \frac{6}{5}\left(\frac{1}{s^2+4} - \frac{1}{s^2+9}\right)$

Step 4: Invert each piece using $\mathcal{L}^{-1}\left\{\frac{\omega}{s^2+\omega^2}\right\} = \sin(\omega t)$:

$y(t) = \cos(2t) + \tfrac{6}{5}\cdot\tfrac{1}{2}\sin(2t) - \tfrac{6}{5}\cdot\tfrac{1}{3}\sin(3t)$

$y(t) = \cos(2t) + \tfrac{3}{5}\sin(2t) - \tfrac{2}{5}\sin(3t)$

Solving Differential Equations with Discontinuous Forcing

Representing Discontinuous Functions

Many real-world inputs switch on or off abruptly. The unit step function (also called the Heaviside function) is the building block for representing these:

$u(t-a) = \begin{cases} 0 & t < a \\ 1 & t \geq a \end{cases}$

Its Laplace transform is $\mathcal{L}\{u(t-a)\} = \frac{e^{-as}}{s}$.

You can build piecewise functions from combinations of step functions. For example, suppose:

$f(t) = \begin{cases} t & 0 \leq t < 2 \\ 4 & t \geq 2 \end{cases}$

Rewrite this as $f(t) = t + (4 - t)\,u(t-2)$. The idea is that $u(t-2)$ "turns on" at $t = 2$ and the term $(4 - t)$ corrects the function from $t$ to $4$ at that moment.

A related tool is the second shifting theorem: if $\mathcal{L}\{f(t)\} = F(s)$, then

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

This lets you transform shifted, switched-on functions directly.

Laplace Transform Method for Discontinuous Forcing

The real advantage here is that you don't need to solve the DE separately on each interval and then match conditions at the breakpoints. The step functions and exponentials in $s$ handle everything in one pass.

Step 1: Transform both sides. With zero initial conditions, the left side simplifies:

$(s^2 + 2s + 1)Y(s) = \frac{e^{-\pi s}}{s}$

Step 2: Solve for $Y(s)$. Note that $s^2 + 2s + 1 = (s+1)^2$:

$Y(s) = \frac{e^{-\pi s}}{s(s+1)^2}$

Step 3: Decompose $\frac{1}{s(s+1)^2}$ with partial fractions:

$\frac{1}{s(s+1)^2} = \frac{1}{s} - \frac{1}{s+1} - \frac{1}{(s+1)^2}$

Step 4: Invert. Without the exponential factor, $\mathcal{L}^{-1}\left\{\frac{1}{s(s+1)^2}\right\} = 1 - e^{-t} - te^{-t}$. The factor $e^{-\pi s}$ shifts everything by $\pi$ and multiplies by $u(t-\pi)$:

$y(t) = \left[1 - e^{-(t-\pi)} - (t-\pi)e^{-(t-\pi)}\right]u(t-\pi)$

The solution is identically zero for $t < \pi$, then responds once the step input kicks in.

Particular Solutions to Non-homogeneous Equations

Laplace Transform Approach

For a non-homogeneous equation like $y'' + ay' + by = g(t)$, the Laplace transform method finds the complete solution (homogeneous + particular) in one shot, since initial conditions are built in.

Step 1: Transform:

$s^2Y(s) - s + 4Y(s) = \frac{2}{s+1}$

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

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

Step 3: Partial fractions on the second term:

$\frac{2}{(s+1)(s^2+4)} = \frac{2}{5}\cdot\frac{1}{s+1} - \frac{2}{5}\cdot\frac{s}{s^2+4} + \frac{2}{5}\cdot\frac{1}{s^2+4}$

Step 4: Invert:

$y(t) = \cos(2t) + \tfrac{2}{5}e^{-t} - \tfrac{2}{5}\cos(2t) + \tfrac{1}{5}\sin(2t)$

$y(t) = \tfrac{3}{5}\cos(2t) + \tfrac{1}{5}\sin(2t) + \tfrac{2}{5}e^{-t}$

You can verify: as $t \to \infty$, the exponential term decays and only the oscillatory part remains. The $e^{-t}$ piece is the transient response to the forcing, while the trig terms come from both the initial conditions and the system's natural frequency.

Comparison with Classical Methods

For certain forcing functions (polynomials, exponentials, sines/cosines), the method of undetermined coefficients also works. For example, to find a particular solution to $y'' + y = t^2$:

1. Guess the form $y_p = At^2 + Bt + C$.
2. Compute $y_p'' = 2A$.
3. Substitute: $2A + At^2 + Bt + C = t^2$.
4. Match coefficients: $A = 1$, $B = 0$, $2A + C = 0 \Rightarrow C = -2$.
5. Result: $y_p = t^2 - 2$.

The Laplace method is generally more efficient when you have initial conditions, discontinuous forcing, or complicated right-hand sides. Undetermined coefficients can be quicker for simple forcing functions when you only need the particular solution form.

Interpreting Solutions in Context

Physical Interpretations

Solutions to these DEs model real systems: spring-mass-damper mechanisms, RLC circuits, control systems, and more. The structure of $y(t)$ tells you about the system's behavior:

• Transient terms (like $e^{-\zeta\omega_n t}\cos(\omega_d t + \phi)$) decay over time and represent the system settling from its initial state. The damping ratio $\zeta$ controls how fast they fade.
• Steady-state terms persist and represent the system's long-term response to the forcing function. For a constant force on a spring, this is just the static displacement $F/k$.
• Poles of $Y(s)$ in the s-domain reveal stability. If all poles have negative real parts, the system is stable (transients decay). Poles with positive real parts mean the response grows without bound.

The final value theorem gives a quick check on long-term behavior: $\lim_{t\to\infty} y(t) = \lim_{s\to 0} sY(s)$, provided the limit exists (all poles of $sY(s)$ in the left half-plane).

Analyzing Discontinuous Solutions

When the forcing function has discontinuities, the solution typically shows distinct behavior before and after each switch point. For a step input that turns on at $t = a$:

• For $t < a$, the system responds only to initial conditions (free response).
• For $t \geq a$, the forced response begins and eventually dominates.

A practical example: in a temperature control system, a thermostat switching on at $t = a$ acts like a step input. The solution shows the system at ambient temperature before $t = a$, then a transient rise toward the target temperature afterward. The exponential terms in the shifted solution $u(t-a)[\cdots]$ capture exactly how quickly the system approaches its new equilibrium.