Solving Differential Equations With Laplace Transform

Solving differential equations can often be a daunting task, but the Laplace transform offers a powerful and elegant method to tackle these problems. This technique transforms differential equations into algebraic equations, which are generally easier to solve. Once the algebraic solution is obtained, an inverse Laplace transform is applied to return to the original domain and find the solution to the differential equation. This approach is particularly useful for linear differential equations with constant coefficients, especially those encountered in engineering, physics, and applied mathematics.

Introduction to Laplace Transforms

The Laplace transform is an integral transform that converts a function of time, t, into a function of complex frequency, s. Mathematically, the Laplace transform of a function f(t), defined for t ≥ 0, is given by:

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

Where:

• F(s) is the Laplace transform of f(t)
• s is a complex frequency parameter
• The integral is evaluated from 0 to infinity

The inverse Laplace transform, denoted as (\mathcal{L}^{-1}{F(s)}), recovers the original function f(t) from its Laplace transform F(s).

Key Properties of Laplace Transforms

Several properties make Laplace transforms particularly useful for solving differential equations:

• Linearity: The Laplace transform of a linear combination of functions is the linear combination of their individual Laplace transforms:

  $\mathcal{L}{af(t) + bg(t)} = a\mathcal{L}{f(t)} + b\mathcal{L}{g(t)} = aF(s) + bG(s)$

• Transform of Derivatives: The Laplace transform of the derivative of a function is:

  $\mathcal{L}{f'(t)} = sF(s) - f(0)$

  For higher-order derivatives:

  $\mathcal{L}{f''(t)} = s^2F(s) - sf(0) - f'(0)$

  $\mathcal{L}{f'''(t)} = s^3F(s) - s^2f(0) - sf'(0) - f''(0)$

  And in general:

  $\mathcal{L}{f^{(n)}(t)} = s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0)$

• Transform of Integrals: The Laplace transform of the integral of a function is:

  $\mathcal{L}\left{\int_0^t f(\tau) , d\tau\right} = \frac{F(s)}{s}$

• Time Shifting (Translation): If (\mathcal{L}{f(t)} = F(s)), then:

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

  Where u(t-a) is the Heaviside step function.

• Frequency Shifting: If (\mathcal{L}{f(t)} = F(s)), then:

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

• Scaling: If (\mathcal{L}{f(t)} = F(s)), then:

  $\mathcal{L}{f(at)} = \frac{1}{a}F\left(\frac{s}{a}\right)$

Common Laplace Transforms

Here are some common functions and their Laplace transforms:

Steps to Solve Differential Equations Using Laplace Transforms

The general process involves the following steps:

1. Transform the Differential Equation: Apply the Laplace transform to both sides of the differential equation. Use the properties of Laplace transforms to convert derivatives into algebraic expressions involving s and the initial conditions.

2. Solve for the Laplace Transform of the Solution: Manipulate the resulting algebraic equation to solve for Y(s), where Y(s) is the Laplace transform of the solution y(t).

3. Perform Partial Fraction Decomposition (if necessary): Decompose Y(s) into simpler fractions that can be easily inverted using standard Laplace transform tables.

4. Apply the Inverse Laplace Transform: Use the inverse Laplace transform to find y(t) from Y(s). This yields the solution to the original differential equation.

Detailed Examples

Let’s illustrate this process with several examples.

Example 1: First-Order Linear Differential Equation

Solve the following first-order linear differential equation:

$y'(t) + 2y(t) = e^{-t}, \quad y(0) = 1$

1. Apply the Laplace Transform: Using the linearity property and the transform of derivatives, we have:

   $\mathcal{L}{y'(t) + 2y(t)} = \mathcal{L}{e^{-t}}$

   $sY(s) - y(0) + 2Y(s) = \frac{1}{s+1}$

2. Substitute Initial Condition and Solve for Y(s): Given y(0) = 1, substitute this into the equation:

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

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

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

3. Partial Fraction Decomposition: Decompose the second term into partial fractions:

   $\frac{1}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}$

   $1 = A(s+2) + B(s+1)$

   Setting s = -1: $1 = A(-1+2) \Rightarrow A = 1$

   Setting s = -2: $1 = B(-2+1) \Rightarrow B = -1$

   So,

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

   Therefore,

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

4. Apply the Inverse Laplace Transform:

   $y(t) = \mathcal{L}^{-1}{Y(s)} = \mathcal{L}^{-1}\left{\frac{1}{s+1}\right}$

   $y(t) = e^{-t}$

   Thus, the solution to the differential equation is y(t) = e^(-t).

Example 2: Second-Order Linear Differential Equation

Solve the following second-order linear differential equation:

$y''(t) + 3y'(t) + 2y(t) = 4e^{t}, \quad y(0) = 1, \quad y'(0) = 0$

1. Apply the Laplace Transform:

   $\mathcal{L}{y''(t) + 3y'(t) + 2y(t)} = \mathcal{L}{4e^{t}}$

   $s^2Y(s) - sy(0) - y'(0) + 3(sY(s) - y(0)) + 2Y(s) = \frac{4}{s-1}$

2. Substitute Initial Conditions and Solve for Y(s):

   $s^2Y(s) - s(1) - 0 + 3(sY(s) - 1) + 2Y(s) = \frac{4}{s-1}$

   $(s^2 + 3s + 2)Y(s) - s - 3 = \frac{4}{s-1}$

   $(s^2 + 3s + 2)Y(s) = s + 3 + \frac{4}{s-1}$

   $Y(s) = \frac{s+3}{s^2 + 3s + 2} + \frac{4}{(s-1)(s^2 + 3s + 2)}$

   $Y(s) = \frac{s+3}{(s+1)(s+2)} + \frac{4}{(s-1)(s+1)(s+2)}$

3. Partial Fraction Decomposition:

   First term: $\frac{s+3}{(s+1)(s+2)} = \frac{A}{s+1} + \frac{B}{s+2}$

   $s+3 = A(s+2) + B(s+1)$

   Setting s = -1: $-1+3 = A(-1+2) \Rightarrow A = 2$

   Setting s = -2: $-2+3 = B(-2+1) \Rightarrow B = -1$

   So,

   $\frac{s+3}{(s+1)(s+2)} = \frac{2}{s+1} - \frac{1}{s+2}$

   Second term: $\frac{4}{(s-1)(s+1)(s+2)} = \frac{C}{s-1} + \frac{D}{s+1} + \frac{E}{s+2}$

   $4 = C(s+1)(s+2) + D(s-1)(s+2) + E(s-1)(s+1)$

   Setting s = 1: $4 = C(2)(3) \Rightarrow C = \frac{2}{3}$

   Setting s = -1: $4 = D(-2)(1) \Rightarrow D = -2$

   Setting s = -2: $4 = E(-3)(-1) \Rightarrow E = \frac{4}{3}$

   So,

   $\frac{4}{(s-1)(s+1)(s+2)} = \frac{2/3}{s-1} - \frac{2}{s+1} + \frac{4/3}{s+2}$

   Therefore,

   $Y(s) = \frac{2}{s+1} - \frac{1}{s+2} + \frac{2/3}{s-1} - \frac{2}{s+1} + \frac{4/3}{s+2}$

   $Y(s) = \frac{2/3}{s-1} + \frac{1/3}{s+2}$

4. Apply the Inverse Laplace Transform:

   $y(t) = \mathcal{L}^{-1}{Y(s)} = \mathcal{L}^{-1}\left{\frac{2/3}{s-1} + \frac{1/3}{s+2}\right}$

   $y(t) = \frac{2}{3}e^{t} + \frac{1}{3}e^{-2t}$

   Thus, the solution to the differential equation is y(t) = (2/3)e^(t) + (1/3)e^(-2t).

Example 3: System of Differential Equations

Consider the system of differential equations:

$x'(t) + y(t) = 0$ $y'(t) + x(t) = 0$

With initial conditions x(0) = 1 and y(0) = 0.

1. Apply the Laplace Transform:

   $\mathcal{L}{x'(t) + y(t)} = 0 \Rightarrow sX(s) - x(0) + Y(s) = 0$ $\mathcal{L}{y'(t) + x(t)} = 0 \Rightarrow sY(s) - y(0) + X(s) = 0$

2. Substitute Initial Conditions:

   $sX(s) - 1 + Y(s) = 0$ $sY(s) + X(s) = 0$

3. Solve the System of Equations:

   From the second equation: $X(s) = -sY(s)$

   Substitute into the first equation: $s(-sY(s)) - 1 + Y(s) = 0$ $-s^2Y(s) + Y(s) = 1$ $Y(s)(1 - s^2) = 1$ $Y(s) = \frac{1}{1 - s^2} = \frac{-1}{s^2 - 1}$

   Then, $X(s) = -sY(s) = \frac{s}{s^2 - 1}$

4. Apply the Inverse Laplace Transform:

   $x(t) = \mathcal{L}^{-1}{X(s)} = \mathcal{L}^{-1}\left{\frac{s}{s^2 - 1}\right} = \cosh(t)$ $y(t) = \mathcal{L}^{-1}{Y(s)} = \mathcal{L}^{-1}\left{\frac{-1}{s^2 - 1}\right} = -\sinh(t)$

   Thus, the solution is x(t) = cosh(t) and y(t) = -sinh(t).

Example 4: Differential Equation with Discontinuous Forcing Function

Solve the following differential equation:

$y''(t) + 4y(t) = f(t), \quad y(0) = 0, \quad y'(0) = 0$

Where

$f(t) = \begin{cases} 1, & 0 \leq t < 2 \ 0, & t \geq 2 \end{cases}$

1. Express f(t) using Heaviside Functions:

   f(t) can be written as:

   $f(t) = 1 - u(t-2)$

2. Apply the Laplace Transform:

   $\mathcal{L}{y''(t) + 4y(t)} = \mathcal{L}{1 - u(t-2)}$

   $s^2Y(s) - sy(0) - y'(0) + 4Y(s) = \frac{1}{s} - \frac{e^{-2s}}{s}$

3. Substitute Initial Conditions and Solve for Y(s):

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

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

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

4. Partial Fraction Decomposition:

   $\frac{1}{s(s^2 + 4)} = \frac{A}{s} + \frac{Bs + C}{s^2 + 4}$

   $1 = A(s^2 + 4) + (Bs + C)s$

   $1 = A(s^2 + 4) + Bs^2 + Cs$

   Comparing coefficients:

   • s²: A + B = 0
   • s: C = 0
   • Constant: 4A = 1 => A = 1/4

   Thus, B = -1/4.

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

5. Apply the Inverse Laplace Transform:

   $y(t) = \mathcal{L}^{-1}\left{\frac{1/4}{s} - \frac{(1/4)s}{s^2 + 4} - e^{-2s}\left(\frac{1/4}{s} - \frac{(1/4)s}{s^2 + 4}\right)\right}$

   $y(t) = \frac{1}{4}\mathcal{L}^{-1}\left{\frac{1}{s}\right} - \frac{1}{4}\mathcal{L}^{-1}\left{\frac{s}{s^2 + 4}\right} - \frac{1}{4}\mathcal{L}^{-1}\left{e^{-2s}\frac{1}{s}\right} + \frac{1}{4}\mathcal{L}^{-1}\left{e^{-2s}\frac{s}{s^2 + 4}\right}$

   $y(t) = \frac{1}{4}(1) - \frac{1}{4}\cos(2t) - \frac{1}{4}u(t-2) + \frac{1}{4}\cos(2(t-2))u(t-2)$

   $y(t) = \frac{1}{4}(1 - \cos(2t)) - \frac{1}{4}u(t-2)(1 - \cos(2(t-2)))$

   Thus, the solution is:

   $y(t) = \begin{cases} \frac{1}{4}(1 - \cos(2t)), & 0 \leq t < 2 \ \frac{1}{4}(\cos(2(t-2)) - \cos(2t)), & t \geq 2 \end{cases}$

Advantages and Limitations

Advantages:

• Simplification: Converts differential equations into algebraic equations, which are easier to solve.
• Handles Initial Conditions Directly: Incorporates initial conditions directly into the transformation, simplifying the solution process.
• Handles Discontinuous Functions: Can handle discontinuous forcing functions and impulse functions effectively using Heaviside step functions and the Dirac delta function.
• Systematic Approach: Provides a systematic approach to solving linear differential equations, reducing the need for ad-hoc methods.

Limitations:

• Limited to Linear Equations: The Laplace transform method is primarily applicable to linear differential equations with constant coefficients. It is not generally suitable for nonlinear differential equations.
• Requires Transforms to Exist: The Laplace transform and its inverse must exist for the functions involved. This requires certain conditions on the functions, such as exponential order.
• Partial Fraction Decomposition: The process of partial fraction decomposition can be complex and time-consuming for higher-order equations.
• Not Always the Most Efficient Method: For some simple differential equations, other methods, such as direct integration or using integrating factors, may be more efficient.

Advanced Techniques and Applications

Convolution Theorem

The convolution theorem states that the Laplace transform of the convolution of two functions is the product of their Laplace transforms:

$\mathcal{L}{(f * g)(t)} = F(s)G(s)$

Where the convolution f * g is defined as:

$(f * g)(t) = \int_0^t f(\tau)g(t - \tau) , d\tau$

This theorem is useful for solving differential equations where the forcing function is given as a convolution integral.

Transfer Functions

In engineering, particularly in control systems, the Laplace transform is used to define transfer functions. The transfer function H(s) of a linear time-invariant (LTI) system is the ratio of the Laplace transform of the output Y(s) to the Laplace transform of the input X(s), assuming zero initial conditions:

$H(s) = \frac{Y(s)}{X(s)}$

Transfer functions are used to analyze the stability and performance of systems, and to design controllers that meet specific performance requirements.

Applications in Electrical Engineering

In electrical engineering, Laplace transforms are used to analyze circuits, especially those involving inductors and capacitors. By transforming the circuit equations into the s-domain, engineers can analyze the circuit's behavior in terms of its impedance and frequency response. This is particularly useful for designing filters and analyzing the stability of feedback amplifiers.

Applications in Mechanical Engineering

In mechanical engineering, Laplace transforms are used to analyze vibrations, control systems, and dynamic systems. For example, the response of a spring-mass-damper system to an external force can be analyzed using Laplace transforms to determine the system's natural frequencies, damping ratio, and stability.

Conclusion

Solving differential equations using Laplace transforms is a powerful technique that simplifies the process by converting differential equations into algebraic equations. While it has limitations, particularly with nonlinear equations, its systematic approach and ability to handle initial conditions and discontinuous functions make it an invaluable tool in various fields, including engineering, physics, and applied mathematics. By understanding the properties of Laplace transforms and mastering the techniques of partial fraction decomposition and inverse Laplace transforms, one can effectively solve a wide range of differential equations.