You have enemies? Good. That means you’ve stood up for something, sometime in your life, Winston Churchill

We shall fight on the beaches, we shall fight on the landing grounds, we shall fight in the fields and in the streets, we shall fight in the hills; we shall never surrender, Winston Churchill

An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:

**Dependent and independent variables**. Variables represent unknown quantities. The independent variable is chosen freely, while the dependent variable changes in response to the independent variable.**Constants**. Fixed numerical values that do not change.**Algebraic operations**. Operations such as addition, subtraction, multiplication, division, exponentiation, and root extraction.

Definition. A differential equation is an equation that involves one or more dependent variables, their derivatives with respect to one or more independent variables, and the independent variables themselves, e.g., $\frac{dy}{dx} = 3x +5y, y’ + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

It involves (e.g., $\frac{dy}{dx} = 3x +5y$):

**Dependent variables**:*Variables that depend on one or more other variables*(y).**Independent variables**: Variables upon which the dependent variables depend (x).**Derivatives**: Rates at which the dependent variables change with respect to the independent variables, $\frac{dy}{dx}$

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:

- The function f(x, y) (the right-hand side of the ODE) in y’ = f(x, y) is continuous in a neighborhood around a point (x
_{0}, y_{0}) and - Its partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous near (x
_{0}, y_{0}).

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x_{0}, y_{0}) .

A first-order linear differential equation (ODE) has the general form: a(x)y' + b(x)y = c(x) where y′ is the derivative of y with respect to x, and a(x), b(x), and c(x) are functions of x. If c(x) = 0, the equation is called homogeneous, i.e., a(x)y’ + b(x)y = 0.

The equation can also be written in the standard linear form as: y’ + p(x)y = q(x) where $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$

In mathematics, a linear ordinary differential equation (ODE) is an equation that involves an unknown function and its derivatives. This property of linearity makes linear ODEs fundamental in modeling various physical systems, including:

- Fluid Dynamics: Describing the flow of fluids under various conditions.
- Electrical Circuits: Analyzing the behavior of circuits involving resistors, capacitors, and inductors.
- Mechanical Vibrations: Modeling the motion of oscillating systems, such as springs and pendulums.
- Radioactive Decay: Describing the rate of decay of radioactive substances.

The linearity of these equations allows for more tractable analytical solutions, often serving as approximations to more complex nonlinear systems. This makes them essential tools in both theoretical and applied mathematics.

The linearity of these ODEs means that the unknown function and its derivatives appear to the first power and are not multiplied together. In this article, we will explore different forms of linear ODEs, their applications, and their solutions, with a focus on understanding their behavior in real-world contexts.

An RC circuit is a fundamental electrical circuit consisting of a resistor (R) and a capacitor (C) connected in series or parallel with a voltage source. Understanding the behavior of an RC circuit is crucial in fields like electrical engineering and physics, as it illustrates how circuits respond over time to changes in voltage.

**Resistor (R)**: A component that opposes the flow of electric current, measured in ohms (Ω).**Capacitor (C)**: A device that stores electrical energy in an electric field, measured in farads (F).**Voltage Source (ε(t))**: It is a component that provides electrical energy to a circuit. It can be constant (like a battery) or time-dependent (like an AC power supply).

In an RC circuit, the relationship between the charge q(t) on the capacitor, the current i(t), and the applied voltage ε(t) is governed by the following first-order linear ordinary differential equation (ODE): $R\frac{dq}{dt}+\frac{q}{C}=ε(t)$ where:

- q is the charge on the capacitor. This represents the electric charge stored in the capacitor at time t.
- $\frac{dq}{dt}$ is the rate of change of charge with respect to time. This is equivalent to the current flowing through the circuit, defined as $i(t) = \frac{dq}{dt}$.
- R is the resistance. This component resists the flow of electric current, measured in ohms (Ω).
- C is the capacitance. This defines the ability of a capacitor to store charge, measured in farads (F). A capacitor is like a tiny rechargeable battery that stores electrical energy in an electric field. Later, this stored charge can be released to do work, like powering a circuit or smoothing out voltage fluctuations.
- ε(t) is the voltage supplied by the voltage source, which can be constant or time-dependent. (Refer to Figure F for a visual representation and aid in understanding it).

How do we arrive to the previous differential equation?

**Ohm’s Law**: The relationship between voltage (V), current (i), and resistance (R) is defined by Ohm’s Law: V = i(t)⋅R = $R·\frac{dq}{dt}$.

**Voltage Across Capacitor**: The voltage across the capacitor is defined as: $V_C = \frac{q(t)}{C}$.

**Kirchhoff’s Voltage Law (KVL)**: The sum of the voltage drops in a closed loop equals the supplied voltage: $V_R + V_C = ε(t)$

$V_R + V_C = ε(t)$ ↭[Substitute V_{R} (Ohm’s Law) and V_{C} (Voltage Across Capacitor):] $R·\frac{dq}{dt} + \frac{q(t)}{C} = ε(t)$

To solve the differential equation, we rewrite it in standard linear form: $\frac{dq}{dt}+\frac{1}{RC}q = \frac{ε(t)}{R}$.

Using the notation q’ = $\frac{dq}{dt},\text{ we have } q’ +\frac{1}{RC}q = \frac{ε(t)}{R}$

This equation is a first-order linear ordinary differential equation, which can be solved using an integrating factor.

- Identify P(t) and Q(t). The standard form of a first-order linear ODE is: y′+P(t)y = Q(t). For our equation: y = q, P(t) =
^{1}⁄_{RC}(a constant), Q(t) =^{ε(t)}⁄_{R}. - Calculate the Integrating Factor μ(t): $μ(t) = e^{\int P(t)dt} = e^{\int \frac{1}{RC}dt} = e^{\frac{t}{RC}}$
- Multiply Both Sides by μ(t): $e^{\frac{t}{RC}}q’ +e^{\frac{t}{RC}}(\frac{1}{RC}q) = e^{\frac{t}{RC}}·\frac{ε(t)}{R}$. Simplify the left side using the product rule: $\frac{d}{dt}(e^{\frac{t}{RC}}·q) = e^{\frac{t}{RC}}·\frac{ε(t)}{R}$
- Integrate Both Sides: $e^{\frac{t}{RC}}·q = \int e^{\frac{t}{RC}}·\frac{ε(t)}{R}dt + C’$ where C’ is the constant of integration.
- Solve for $q(t) = e^{\frac{-t}{RC}}(\int e^{\frac{t}{RC}}·\frac{ε(t)}{R}dt + C’)$

The solution q(t) consists of two parts:

**Homogeneous Solution (Natural Response)**: $q_h(t) = C’e^{\frac{-t}{RC}}$. Represents the transient behavior, which decays exponentially over time.**Particular Solution (Forced Response)**: Depends on ε(t) and represents the steady-state behavior of the circuit.

When the voltage source is constant:

- Integrate the right side: $\int e^{\frac{t}{RC}}·\frac{ε(t)}{R}dt = \int e^{\frac{t}{RC}}·\frac{E}{R}dt = \frac{E}{R}\int e^{\frac{t}{RC}} dt = \frac{E}{R}·RC·e^{\frac{t}{RC}} = EC·e^{\frac{t}{RC}}$.
- Plug back into the solution: $q(t) = e^{\frac{-t}{RC}}(EC·e^{\frac{t}{RC}} + C’) = EC + C’e^{\frac{-t}{RC}}$
- Determine the Constant C’ Using Initial Conditions q(0) = q
_{0}. q_{0}= EC + C’ ⇒ C’ = q_{0}- EC. $q(t) = EC + (q_0 -EC)e^{\frac{-t}{RC}}$. - Current i(t) in the Circuit. Since i(t) = $\frac{dq}{dt} = \frac{d}{dt}(EC) + \frac{d}{dt}((q_0 -EC)e^{\frac{-t}{RC}}) = 0 +(q_0 -EC)\frac{-1}{RC}e^{\frac{-t}{RC}} = -\frac{q_0 -EC}{RC}e^{\frac{-t}{RC}}$.

Radioactive decay is a natural process by which an unstable atomic nucleus loses energy by emitting radiation. This phenomenon can be mathematically modeled using first-order linear ordinary differential equations (ODEs). In this context, we will explore a system where one radioactive substance A decays into another B, and then that second substance also decays.

Consider two substances:

**Substance A**: A decays into substance B at a rate proportional to the amount of A, A(t). Let k_{1}be the decay constant for substance A.**Substance B**: B decays at a rate proportional to the amount of B, B(t). Let k_{2}be the decay constant for substance B.

The rate of decay of Substance A is proportional to its current amount: $\frac{dA}{dt} = -k_1A$.

This is a separable differential equation, and its solution is: $A = A_0e^{-k_1t}$, where A_{0} is the initial amount of substance A at t = 0.

Substance B gains mass from the decay of A and loses mass due to its own decay. The rate of change of B is given by: $\frac{dB}{dt} = k_1A(t) -k_2B(t)$

Substituting A(t) into the equation: $\frac{dB}{dt} = k_1A_0e^{-k_1t}-k_2B⇒ B’ + k_2B = k_1A_0e^{-k_1t}$. This is a first-order linear ODE of the form y' + P(t)y = Q(t). The solution describes the amount of B as it accumulates and decays over time,

**Find the Integrating Factor**: $μ(t) = e^{\int P(t)dt}= e^{\int k_2dt} = e^{k_2·t}$**Multiply Both Sides by μ(t)**: $e^{k_2·t}\frac{dB}{dt}+ e^{k_2·t}k_2B = k_1A_0e^{-k_1t}e^{k_2·t}↭[Simplify] \frac{d}{dt}(e^{k_2·t}B(t)) = k_1A_0e^{(k_2-k_1)t}$**Integrate Both Sides**: $\int \frac{d}{dt}(e^{k_2·t}B(t)) = \int k_1A_0e^{(k_2-k_1)t} ↭ e^{k_2·t}B(t) = \int k_1A_0e^{(k_2-k_1)t} + C$ where C is the constant of integration.**Evaluate the Integral**: Depending on the values of k_{1}and k_{2}, we have two cases. Case 1: k_{1}≠ k_{2}, $e^{k_2·t}B(t) = \frac{k_1A_0}{k_2-k_1}e^{(k_2-k_1)t} + C$. Case 2: k_{1}= k_{2}, $e^{k_2·t}B(t) = k_1A_0t + C$**Solve for B(t)**.

For k_{1} ≠ k_{2}, $B(t) = e^{-k_2·t}(\frac{k_1A_0}{k_2-k_1}e^{(k_2-k_1)t} + C)↭ B(t) = \frac{k_1A_0}{k_2-k_1}e^{-k_1t} + Ce^{-k_2·t}$

For k_{1} = k_{2} = k, $B(t) = e^{-k·t}(kA_0t + C)$

- Particular solution, $B_p(t) = \frac{k_1A_0}{k_2-k_1}e^{-k_1t}$. Represents the accumulation of B due to the decay of A.
- Homogeneous Solution, $B_h(t) = Ce^{-k_2·t}$. Represents the natural decay of B independent of the substance A.

To determine C, we need an initial condition. If we assume that at time t = 0, the amount of B is B(0) = B_{0}, then: $B(t) = B_0 = \frac{k_1A_0}{k_2-k_1}e^{-k_1·0} + Ce^{-k_2··0}↭ B_0 = \frac{k_1A_0}{k_2-k_1} + C ⇒[\text{Solve for C}] C = B_0 - \frac{k_1A_0}{k_2-k_1}$

$B(t) = \frac{k_1A_0}{k_2-k_1}e^{-k_1t} + (B_0 - \frac{k_1A_0}{k_2-k_1})e^{-k_2·t}$

- Initially, B increases as A decays into it. Over time, B also decays, reducing its amount.
- If k
_{1}> k_{2}, the accumulation of B is more pronounced before it starts to decay significantly. If k_{1}< k_{2}, B decays quickly, and its accumulation is less noticeable. - The graph typically shows an initial rise in B(t), reaching a peak (where the rate of accumulation equals the rate of decay), and then a decline.

This different equation is different from previous cases because of the negative sign before ay. It is a first-order linear ODE and can exhibit exponentially growing solutions.

- It is already in standard form: $\frac{dy}{dt}-ay = q(t)$. To solve it, we use the integrating factor method.
- Find the Integrating Factor: $μ(t) = e^{\int -adt} = e^{-at}$
- Multiply Both Sides by μ(t): $e^{-at}\frac{dy}{dt}-ae^{-at}y = e^{-at}q(t) ↭ \frac{d}{dt}(e^{-at}y) = e^{-at}q(t)$
- Integrate Both Sides: $\int \frac{d}{dt}(e^{-at}y)dt = \int e^{-at}q(t)dt ↭ e^{-at}y = \int e^{-at}q(t)dt + C$
- Solve for y: $y(t) = e^{at}\int q(t)e^{-at}dt + ce^{at}$.

Unlike the earlier transient terms (which decay over time), the term $ce^{at}$ is not transient anymore, but grows exponentially since a > 0.

If C ≠ 0, the homogeneous solution $ce^{at}$ grows without bound. The particular solution depends on q(t) and may or may not counteract the growth.

The constant C is determined by the initial condition y(t_{0}) = y_{0}. Small differences in y_{0} can lead to large differences in y(t) as t increases due to the exponential growth.

The equation becomes: $\frac{dy}{dt}-ay = q_0$.

- Integrate: $\int q(t)e^{-at}dt = \int q_0e^{-at}dt = -\frac{q_0}{a}e^{-at} + C’$ (Since C’ will be absorbed into C, we can omit it).
- Simplify $y(t) = e^{at}(-\frac{q_0}{a}e^{-at} + C) = -\frac{q_0}{a} + Ce^{at}$
- Apply initial condition. If y(0) = y
_{0}, $y_0 = -\frac{q_0}{a} + Ce^{a·0} = -\frac{q_0}{a} + C [\text{Solve for C}] C = y_0 + \frac{q_0}{a}$. - Final solution: $y(t) = -\frac{q_0}{a} + (y_0 + \frac{q_0}{a})e^{at}$

The solution grows exponentially if $y_0 + \frac{q_0}{a} ≠ 0$. The term $-\frac{q_0}{a}$ is a constant steady-state solution. The behavior heavily depends on the initial condition y_{0}.

In this section, we will explore how to solve first-order linear ordinary differential equations (ODEs) with sinusoidal inputs. Specifically, we will delve into the differential equation:x ’ + kx = cos(wt), where:

- k is a positive constant representing the rate of decay or growth,
- w is the angular frequency of the sinusoidal input.

We will employ the method of complex exponentials to find the general and particular solutions, and extend our analysis to handle more complex inputs using the principle of superposition.

To simplify the solution process, we express the cosine function as the real part of a complex exponential. Using Euler’s formula: cos(wt) = Re[e^{iwt}]

By substituting the complex exponential into the original differential equation, we obtain a complex ODE: z’ + kz = e^{iwt}. This equation is easier to solve, and we can later take the real part to find the solution to the original equation.

A first-order linear ODE can be solved using an integrating factor. The standard form is: y′ +P(t)y = Q(t). For our complex ODE, P(t) = k, Q(t) = e^{iwt}.

Integrating factor: $μ(t) = e^{\int kdt} = e^{kt}$

Multiplying both side of the complex ODE by the integrator factor: $z’e^{kt} + kze^{kt} = e^{kt}e^{iwt}↭ \frac{d}{dt}z·e^{kt}= e^{(k + iw)t}$

Integrate both sides with respect to t: $\int \frac{d}{dt}z·e^{kt}dt = \int e^{(k + iw)t}dt ↭ z·e^{kt} = \frac{1}{k+iw}e^{(k + iw)t} + C$ where C is the constant of integration.

Solving for z: $z = ce^{-kt}+ \frac{1}{k+iw}e^{iwt}$. This is **the general solution to the complex ODE**.

This expression consists of two parts:

**Homogeneous Solution**(Transient Solution): The solution that comes from the homogeneous equation, $ce^{-kt}$, it decays exponentially over time.**Particular Solution**(Steady-State Solution). The solution that comes from the sinusoidal input is the particular solution, $\frac{1}{k+iw}e^{iwt}$.

To find the real solution x(t), we take the real part of z(t): x = Re[z] = $ce^{-kt}+ Re[\frac{1}{k+iw}e^{iwt}]$

$\frac{1}{k+iw}e^{iwt} =[\text{Multiply numerator and denominator by the complex conjugate k−iω:}] \frac{1}{k+iw}·\frac{k-iw}{k-iw}e^{iwt} =[\text{Using Euler’s formula}] \frac{k-iw}{k^2+w^2}(cos(wt)+isin(wt))$

x = $ce^{-kt}+ Re[\frac{1}{k+iw}e^{iwt}] = ce^{-kt}+ \frac{1}{k^2+w^2}[kcos(wt)+wsin(wt)]$

This expression consist of two parts:

**Homogeneous Solution (Transient Solution)**: $ce^{-kt}$, decays exponentially over time due to the e^{-kt}term. Represents the natural response of the system.**Particular Solution (Steady-State Solution)**: $\frac{1}{k^2+w^2}[kcos(wt)+wsin(wt)]$ oscillates with the same frequency w as the input. Represents the forced response due to the input cos(ωt).

Let’s consider a more general differential equation where the input is a scaled sine function: x’ + kx = Fsin(wt), where F is a constant amplitude.

Using Euler’s formula: sin(wt) = Im(e^{iwt}). Formulate the complex ODE: z’ + kz = Fe^{iwt}.

Similar analysis, the general solution to the complex ODE is: $z = ce^{-kt}+ \frac{F}{k+iw}e^{iwt}$

$\frac{F}{k+iw}e^{iwt} = \frac{F(k-iw)}{k^2+w^2}(cos(wt)+isin(wt))$

x = $ce^{-kt}+ Im[\frac{F}{k+iw}e^{iwt}] = ce^{-kt}+ \frac{F}{k^2+w^2}[ksin(wt)-wcos(wt)]$

Finally, let’s solve x’ + kx = cos(wt) + 3sin(wt).

The superposition principle states that for linear systems, the response (output) due to multiple inputs is the sum of the responses due to each input individually.

Split the equation into two separate equations: x’_{1} + k_{1} = cos(wt), x’_{2} + kx_{2} = 3sin(wt).

Find the particular solutions for each equation: x_{p1}(t) = $\frac{1}{k^2+w^2}[kcos(wt)+wsin(wt)]$, x_{p2}(t) = $\frac{3}{k^2+w^2}[ksin(wt)-wcos(wt)]$

The overall particular solution is x_{p} = x_{p1} + x_{p2} = $\frac{1}{k^2+w^2}[kcos(wt)+wsin(wt)+3ksin(wt)-3wcos(wt)]$

Write the general solution (include the homogeneous solution): x(t) = $ce^{-kt} + \frac{1}{k^2+w^2}[kcos(wt)+wsin(wt)+3ksin(wt)-3wcos(wt)]$

- $ce^{-kt}$ is the solution to the homogeneous equation x′ + kx = 0.
- The particular solution x
_{p}(t) accounts for the input functions cos(wt) and 3sin(wt).

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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