Comprehensive Study Note: Solving Differential Equations with Laplace Transforms

Definition and Domain Conversion: The Laplace transform is an integral transform that maps a function of time, $f(t)$ , from the time domain ( $t$ -domain) into a function of a complex variable, $F(s)$ , in the frequency domain ( $s$ -domain).
Mathematical Formula: The formal definition of the unilateral Laplace transform is given by the integral: * $L{f(t)} = F(s) = \int_{0}^{\infty} f(t) e^{-st} \,dt$
Variable Definitions: * $t$ : Usually represents time, where $t \geq 0$ . * $s$ : A complex number frequency parameter, defined as $s = \sigma + i\omega$ .
Purpose of the Transform: It is an essential tool in engineering and physics, specifically for solving linear ordinary differential equations (ODEs). It transforms differential operations into algebraic operations, which are significantly easier to manipulate and solve.

Input Force/Constant ( $F_0$ ): The value provided is $F_0 = 400$ . In physical systems, this typically represents a constant magnitude force or a step input applied to the system at $t = 0$ .
Numerical Parameters: The transcript lists several specific values which are often coefficients for a second-order differential equation or a transfer function: * Parameter A: $20$ * Parameter B: $-36$ * Parameter C: $-210$
Hypothetical System Equation: If these values represent a characteristic equation of a system (such as $m y'' + c y' + k y = f(t)$ ), the governing equation based on the transcript values could be modeled as: * $20 y''(t) - 36 y'(t) - 210 y(t) = 400$

First Derivative Rule: To solve differential equations, the transform of the derivative must be known. For a function $y(t)$ , the transform of its first derivative is: * $L{y'(t)} = sY(s) - y(0)$ * Where $y(0)$ is the initial condition of the system at time zero.
Second Derivative Rule: The transform for the second derivative is: * $L{y''(t)} = s^2Y(s) - s y(0) - y'(0)$ * Where $y'(0)$ is the initial velocity or rate of change at time zero.
Linearity Property: The Laplace transform is a linear operator, meaning: * $L{a f(t) + b g(t)} = a L{f(t)} + b L{g(t)}$

Step 1: Applying the Transform: Apply the Laplace transform to every term in the differential equation using the linearity property. * Assuming the equation $20 y''(t) - 36 y'(t) - 210 y(t) = 400$ * $20 [s^2 Y(s) - s y(0) - y'(0)] - 36 [s Y(s) - y(0)] - 210 Y(s) = \frac{400}{s}$
Step 2: Substitution of Initial Conditions: If the system starts from rest, substitute $y(0) = 0$ and $y'(0) = 0$ . * $(20s^2 - 36s - 210) Y(s) = \frac{400}{s}$
Step 3: Solving for $Y(s)$ : Isolate the output variable in the $s$ -domain: * $Y(s) = \frac{400}{s(20s^2 - 36s - 210)}$
Step 4: Partial Fraction Decomposition: Break the complex fraction into simpler terms that correspond to known transform pairs. * Factor the quadratic expression: $20s^2 - 36s - 210$ . * Simplify by dividing by a common factor (e.g., $2$ ): $10s^2 - 18s - 105$ . * Set up the decomposition: $Y(s) = \frac{A}{s} + \frac{B}{s - r_1} + \frac{C}{s - r_2}$ , where $r_1$ and $r_2$ are roots of the quadratic.
Step 5: Inverse Laplace Transform: Use the inverse transform ( $L^{-1}$ ) to return to the time domain. * $y(t) = L^{-1}{Y(s)}$

Constant (Step) Function: $L{a} = \frac{a}{s}$
Exponential Function: $L{e^{at}} = \frac{1}{s - a}$
Polynomial Function ( $n$ is a positive integer): $L{t^n} = \frac{n!}{s^{n+1}}$
Sine Function: $L{\sin(at)} = \frac{a}{s^2 + a^2}$
Cosine Function: $L{\cos(at)} = \frac{s}{s^2 + a^2}$
Hyperbolic Sine: $L{\sinh(at)} = \frac{a}{s^2 - a^2}$
Hyperbolic Cosine: $L{\cosh(at)} = \frac{s}{s^2 - a^2}$