THE CLO2 was comprised of these 4 major topics:
Module 3: Applications of 1st Order D.E.
3.1. Decomposition / Growth
3.2. Newton’s Law of Cooling
3.3. Mixing (Non-Reacting Fluids)
Module 4: Linear Differential Equation of Order n
4.1. Introduction
4.1.1. Standard form of a nth order Linear DE
4.1.2. Differential Operators
4.1.3. Principle of Superposition
4.1.4. Linear Independence of a Set of Functions
4.2. Homogeneous Linear Differential Equation with Constant Coefficients
4.2.1. Solution of a Homogeneous Linear Ordinary DE
4.2.2. Initial and Boundary Value Problems
Module 5: Laplace Transforms of Functions
5.1. Definition
5.2. Transform of Elementary Functions
5.3. Transform of e at f(t) – Theorem
5.4. Transform of t n f(t) – Derivatives of Transforms
5.5. Inverse Transforms
5.7. Transforms of Derivatives
5.8. Initial Value Problems
Module 6: The Heaviside Unit-Step
6.1. Definition
6.2. Laplace Transforms of Discontinuous Functions and Inverse Transform Leading to Discontinuous Functions
6.3. Solution of Initial Value Problems with Discontinuous Functions by Laplace Transform Method
Module 3: Application of 1st Order D.E.
One general application that is reflected in mine in this module was about the decomposition/growth. To further discuss, this is one of the main topics of this module.
Growth and Decay
If a quantity y is a function of time t and is directly proportional to its rate of change (y’), then we can express the simplest differential equation of growth or decay.
y’ ∝ y
y’ = ky, where k is the constant of proportionality
Solving this DE using separation of variables and expressing the solution in its exponential form would lead us to: y = Cekt
This simple general solution consists of the following: (1) C = initial value, (2) k = constant of proportionality, and (3) t = time. If k > 0, then it is a growth model. Otherwise, if k < 0, then it is a decay model.
One problem application that was stuck in my mind in this problem set would be item #1, the application tackles about the decomposition/growth. To expound, the formula states that y = Cekt. Upon substituting the given values, we can arrive to the final answer which is approximately 11.7 million pesos.
In finality, I do believe those differential equations are the basis of correctness in different situations concerning the world. This subject is created not to make us suffer, but to make us learn better.
P.S. I only incorporated module 3 to the CLO2 since the module talks about the application of topics to selected engineering problems.
It is evident in this topic that I applied differential equations to selected engineering problems.