Introduction to First Order Differential Equations
First-order differential equations contain only the first derivative of the function and can be expressed in the general form:
dy/dt + P(t)y = Q(t)
Here, P(t) and Q(t) are functions of the independent variable t. These equations are pivotal in describing various physical phenomena and can be solved using several methods, depending on their type and complexity.
Understanding the fundamentals of differentiation is crucial for solving these equations.
Types of First Order Differential Equations
- Separable Equations: These equations can be separated into two functions, one solely involving y and the other only t.Example: dy/dt = yt
- Homogeneous Equations: These are equations where every term is a function of y/t or t/y. For a deeper understanding, studying integration of trigonometric functions can provide additional insights into solving these types of equations.Example: ty' - y = (t2 + 1)(y/t)
Applications in Growth and Decay
Exponential Growth and Decay
Exponential growth and decay are phenomena that can be modelled using first-order differential equations. The general form of such equations is:
dy/dt = ky
Where:
- y represents the quantity that is increasing or decreasing over time t,
- k is the proportionality constant.
This model is closely related to exponential functions and their properties.
Example 1: Population Growth
Consider a population of bacteria that doubles every hour. If the initial population is 100, find the population after 3 hours.
Solution: Using the general solution y(t) = y(0)e(kt), where y(0) is the initial population, and given that the population doubles every hour (k = ln(2)), we find:
y(3) = 100 * e(3 * ln(2))
Calculating this gives us the population after 3 hours.
Calculating this gives us the population after 3 hours. To fully grasp the concept of e, refer to the study on logarithmic functions.
Radioactive Decay
Radioactive decay follows a similar model to exponential decay, with the general form:
dN/dt = -λN
Where:
- N is the quantity of the substance that remains after time t,
- λ is the decay constant.
Example 2: Radioactive Decay
Suppose a sample of a radioactive substance decays to 80% of its original amount in 5 days. Find the half-life of the substance.
Solution: Using the decay equation N(t) = N(0)e(-λt), and given data, we can find λ and subsequently use the half-life formula T(1/2) = ln(2)/λ to find the half-life.
Applications in Electrical Circuits
RC Circuits
In electrical circuits, particularly in RC (resistor-capacitor) circuits, first-order differential equations can be used to describe the charging and discharging of a capacitor over time. These applications offer a practical glimpse into the world of second-order differential equations.
Charging a Capacitor
The voltage across the capacitor as a function of time, V(t), while it is being charged, can be described by:
dV/dt = (V0 - V)/(RC)
Where:
- V0 is the source voltage,
- R is the resistance,
- C is the capacitance.
Example 3: Charging a Capacitor
If a capacitor takes 2 seconds to charge up to 60% of the source voltage, find the time constant of the circuit.
Solution: The time constant τ is given by τ = RC. Using the charging equation V(t) = V0 * (1 - e(-t/RC)), we can substitute the given values to find RC. The time constant of the circuit is approximately 2.18 seconds.
Discharging a Capacitor
When discharging, the voltage across the capacitor as a function of time, V(t), is given by:
dV/dt = -V/(RC)
Example 4: Discharging a Capacitor
Given an RC circuit with R = 10 Ω and C = 100 μF, find the voltage across the capacitor 2 seconds after discharging begins.
Solution: Using the discharging equation V(t) = V0 * e(-t/RC), we can substitute the given values to find V(2). This calculation ties into understanding the basics of RC circuits and their behaviour over time, which is a practical application of first-order differential equations.
Through the integration of these links, the study notes not only provide foundational knowledge on first-order differential equations but also encourage a deeper exploration into related mathematical concepts that are essential for a comprehensive understanding of the subject matter.
FAQ
The Bernoulli differential equation is a type of first-order differential equation and is expressed as dy/dt + P(t)y = Q(t)yn, where n is a real number. It is related to standard first-order differential equations but involves a term that is a power of the dependent variable, y. To solve a Bernoulli differential equation, a substitution, such as v = y(1-n), is made to transform it into a linear first-order differential equation, which can then be solved using methods like the integrating factor. Understanding Bernoulli differential equations is essential as they appear in various physical models, including certain types of fluid flow and exponential growth and decay problems.
The existence and uniqueness theorem states that, given a first-order ordinary differential equation (ODE) in the form dy/dt = f(t, y) with an initial condition y(t0) = y0, there exists a unique solution to the ODE if f and ∂f/∂y are continuous in a region of the (t, y)-plane containing the point (t0, y0). This theorem is crucial in the study of differential equations as it assures us that under certain conditions, a unique solution exists, providing a deterministic prediction of the system's future behaviour. It ensures that the solutions derived from differential equations are reliable and consistent in modelling real-world phenomena.
The integrating factor is used to transform a non-exact first-order linear differential equation into an exact equation that can be easily integrated. Given a differential equation in the form dy/dt + P(t)y = Q(t), the integrating factor, μ(t), is e(∫P(t)dt). Multiplying the entire differential equation by μ(t) makes it take the form d(yμ)/dt = μQ, which can be directly integrated to find the solution. The integrating factor is derived by seeking a function that, when multiplied with the original equation, allows the left-hand side to be expressed as the derivative of a product of functions, thereby simplifying the integration process and making the solution more straightforward.
A singular solution of a first-order differential equation is a solution that cannot be obtained from the general solution by choosing particular values for the arbitrary constants. In a physical context, a singular solution often represents a boundary or a threshold between different types of behaviour of a system. For example, in population dynamics, a singular solution might represent a critical population level that separates extinction from unbounded growth. Understanding singular solutions is vital as they can provide insights into the inherent properties and critical states of the physical system being modelled, offering valuable information for analysis and predictions.
The general solution of a first-order linear differential equation dy/dt + P(t)y = Q(t) is given by y(t) = e(-∫P(t)dt) * (∫Q(t)e(∫P(t)dt)dt + C), where C is the constant of integration. This solution is crucial because it provides a function, y(t), that satisfies the original differential equation for any value of C, meaning it represents a family of solutions. The particular solution can be found by substituting a specific initial condition into the general solution. Understanding the general solution is vital as it provides insights into the behaviour of dynamic systems described by the differential equation, such as population dynamics, electrical circuits, and more.
Practice Questions
The general formula for exponential growth is given by P(t) = P0 * e(kt), where P(t) is the population at time t, P0 is the initial population, k is the growth constant, and t is time. Given that P(0) = 200 and P(4) = 800, we can substitute these into the equation to find k. Using P(4) = 800, we get 800 = 200 * e(4k). Solving for k, we find that k = ln(4)/4. Now, to find the time when P(t) = 1600, we substitute P(t) and k back into the equation: 1600 = 200 * e(ln(4)t/4). Solving for t, we find that t = 6. Therefore, the population will reach 1600 bacteria 6 hours after 12:00 PM, which is at 6:00 PM.
The voltage across a discharging capacitor in an RC circuit is given by V(t) = V0 * e(-t/RC), where V(t) is the voltage at time t, V0 is the initial voltage, R is the resistance, C is the capacitance, and t is time. Given that V0 = 10V, R = 5Ω, and C = 200μF = 200 * 10(-6) F, we can substitute these values into the equation to find V(2). So, V(2) = 10 * e(-2/(520010^(-6))). Calculating this, we find that V(2) = 10 * e(-2/0.001) = 10 * e(-2000). Therefore, the voltage across the capacitor 2 seconds after the discharging begins is approximately 0V, as the exponential term is practically zero.
Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.