Basic linear ODE II.

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Recall

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, $\frac{dy}{dx}$

First-Order Linear Ordinary Differential Equations (ODEs)

Definition. A first-order linear ordinary differential equation is an ordinary differential equation (ODE) involving an unknown function y(x), its first derivative y′, and functions of the independent variable x, which can be written in the general form:: a(x)y' + b(x)y = c(x) where:

• y = y(x) is the unknown function of the independent variable x.
• y′ = $\frac{dy}{dx}$ is the derivative of y with respect to x.
• a(x), b(x), and c(x) are known functions of x.

These equations are termed “linear” because the unknown function y and its derivative y’ appear to the first power and are not multiplied together or composed in any nonlinear way.

If the function c(x)=0 for all x in the interval of interest, the equation simplifies to: a(x)y’ + b(x)y = 0. Such an equation is called a homogeneous linear differential equation.

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 the equation y’ = f(x, y) is continuous in some neighborhood around a point (x₀, y₀) and
• Its partial derivative with respect to y, denoted as $\frac{∂f}{∂y}$, is also continuous near (x₀, y₀).

Then, the differential equation y' = f(x, y) has a unique solution to the initial value problem through the point (x₀, y₀), meaning that it satisfies the initial condition y(x₀) = y₀.

This theorem ensures that under these conditions, the solution exists and is unique near x = x₀.

Basic linear ODE

In mathematics, a linear ordinary differential equation (ODE) is an equation involving an unknown function and its derivatives, where both the function and its derivatives appear linearly. This means that the unknown function and its derivatives are raised only to the first power and are not multiplied together or composed within nonlinear functions.

This property of linearity makes linear ODEs fundamental in modeling various physical systems due to their simplicity and the powerful mathematical tools available for solving them. They often arise in fields such as:

• Fluid Dynamics: Describing the flow of fluids under various conditions where the flow can be approximated as linear.
• 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, where Hooke’s law (a linear relationship) applies.
• Radioactive Decay: Describing the rate at which unstable nuclei decay over time, which follows a linear differential equation leading to exponential decay.

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.

In this second article, we will continue exploring different forms of linear ODEs, their applications, and their solutions, with a focus on understanding their behavior in real-world contexts.

Electrical Circuit (RC Circuit)

Components of the RC Circuit

• A resistor (R) opposes the flow of electric current, converting electrical energy into heat. It is measured in ohms (Ω). It determines how quickly the capacitor charges or discharges by limiting the current flow.
• A capacitor (C) stores electrical energy in an electric field between its plates. It is measured in farads (F). It accumulates and releases charge over time, affecting the voltage across its terminals.
• A Voltage Source (ε(t)) provides electrical energy to the circuit. It can be constant (like a battery) or time-dependent (like an AC power supply). It drives the current that charges or discharges the capacitor.

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 because it illustrates how circuits respond over time to changes in voltage. This response is characterized by the charging and discharging of the capacitor, which is governed by a first-order linear ordinary differential equation (ODE).

Governing Differential Equation

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): $R\frac{dq}{dt}+\frac{q(t)}{C}=ε(t)$ where:

• q(t) represents the electric charge stored in the capacitor at time t (in columbus, C).
• $\frac{dq}{dt}$ is the rate of change of charge with respect to time. This is equivalent to the current i(t) 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).

Derivation of the Differential Equation

To derive this equation, we apply fundamental circuit laws:

1. Ohm’s Law: The relationship between voltage (V), current (i), and resistance (R) is defined by Ohm’s Law. The voltage drop across the resistor V_R is: V_R = i(t)⋅R = $R·\frac{dq}{dt}$.
2. Voltage Across Capacitor: The voltage across the capacitor is defined as: $V_C = \frac{q(t)}{C}$.
3. Kirchhoff’s Voltage Law (KVL): The sum of the voltage drops around 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):] $R·\frac{dq}{dt} + \frac{q(t)}{C} = ε(t)$

Standard Form of the Differential Equation

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.

Solving the Differential Equation

1. 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.
2. Calculate the Integrating Factor μ(t): $μ(t) = e^{\int P(t)dt} = e^{\int \frac{1}{RC}dt} = e^{\frac{t}{RC}}$
3. 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: $\frac{d}{dt}(e^{\frac{t}{RC}}·q) = e^{\frac{t}{RC}}·\frac{ε(t)}{R}$
4. 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.
5. Solve for $q(t) = e^{\frac{-t}{RC}}(\int e^{\frac{t}{RC}}·\frac{ε(t)}{R}dt + C’)$

Analyzing the Solution

The general solution q(t) consists of two components:

1. Homogeneous Solution (Natural Response): $q_h(t) = C’e^{\frac{-t}{RC}}$. Represents the transient behavior, which decays exponentially over time due to the discharge of the capacitor.
2. Particular Solution (Forced Response): $q_p(t) = e^{\frac{-t}{RC}}\int e^{\frac{t}{RC}}·\frac{ε(t)}{R}dt$. Depends on the form of ε(t) and represents the steady-state behavior as the circuit responds to the voltage source.

Example: Constant Voltage Source ε(t)=E

When the voltage source is constant, ε(t)=E:

1. Integrate the right side: $\int e^{\frac{t}{RC}}·\frac{ε(t)}{R}dt = \int e^{\frac{t}{RC}}·\frac{E}{R}dt = \frac{E}{R}\int e^{\frac{t}{RC}} dt = \frac{E}{R}·RC·e^{\frac{t}{RC}} = EC·e^{\frac{t}{RC}}$.
2. Plug back into the solution: $q(t) = e^{\frac{-t}{RC}}(EC·e^{\frac{t}{RC}} + C’) = EC + C’e^{\frac{-t}{RC}}$
3. Determine the Constant C’ Using Initial Conditions q(0) = q₀. q₀ = EC + C’ ⇒ C’ = q₀ - EC. $q(t) = EC + (q_0 -EC)e^{\frac{-t}{RC}}$.
4. Current i(t) in the Circuit. Since i(t) = $\frac{dq}{dt} = \frac{d}{dt}(EC) + \frac{d}{dt}((q_0 -EC)e^{\frac{-t}{RC}}) = 0 +(q_0 -EC)\frac{-1}{RC}e^{\frac{-t}{RC}} = -\frac{q_0 -EC}{RC}e^{\frac{-t}{RC}}$.

Interpretation

• Steady State Charge: As t → ∞, $e^{\frac{-t}{RC}}→0, q(∞) = EC$. The capacitor charges up to EC, the maximum charge determined by the voltage and capacitance.
• Transient Behavior: The term $(q_0 -EC)e^{\frac{-t}{RC}}$ represents the transient response, which decays exponentially.
• If q₀< EC, q₀ -EC < 0, so i(t) > 0: Current flows into the capacitor (charging). If q₀ > EC, q₀ -EC > 0, so i(t) < 0: Current flows out of the capacitor (discharging). The current decreases exponentially over time, approaching zero as the capacitor reaches its steady-state charge.
• The time constant τ is a key parameter: τ = RC. It determines the rate at which the capacitor charges or discharges. After time τ, the transient term $e^{\frac{-t}{RC}} = e^{-1} ≈0.3679$. Approximately 63% of the charging or discharging process is complete.

Radioactive Decay

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 B also undergoes radioactive decay.

Consider two substances:

• Substance A: A decays into substance B at a rate proportional to the amount of A, A(t). Let k₁ be the decay constant for substance A.
• Substance B: It accumulates due to the decay of A and simultaneously decays on its own. B decays at a rate proportional to the amount of B, B(t). Let k₂ be the decay constant for substance B.

Modeling the Decay of Substance A and B

The rate at which substance A decays is proportional to its current amount. This relationship is expressed by the differential equation: $\frac{dA}{dt} = -k_1A$.

This is a separable differential equation. Separate variables: $\frac{dA}{A} = -k_1dt ↭[\text{Integrate both sides}] \int \frac{dA}{A} = \int -k_1dt ↭ ln|A| = -k_1t + C_1 ↭[\text{Solve for A}] A(t) = e^{-k_1t+C_1} = A_0e^{-k_1t}$ where A₀ = e^C₁ is the initial amount of substance A at t = 0.

The amount of substance A decreases exponentially over time.

Substance B gains mass from the decay of substance A and loses mass due to its own decay. The rate of change of B(t) is the difference between the rate at which A decays into B (A decays at rate $\frac{-dA}{dt} = k_1A(t)$) and the rate at which B decays (proportional to its current amount: k₂B(t)): $\frac{dB}{dt} = k_1A(t) -k_2B(t)$

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,

Solving the Differential Equation for B(t)

1. Find the Integrating Factor: $μ(t) = e^{\int P(t)dt}= e^{\int k_2dt} = e^{k_2·t}$
2. Multiply Both Sides by μ(t): $e^{k_2·t}\frac{dB}{dt}+ e^{k_2·t}k_2B = k_1A_0e^{-k_1t}e^{k_2·t}↭[Simplify] \frac{d}{dt}(e^{k_2·t}B(t)) = k_1A_0e^{(k_2-k_1)t}$
3. 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.
4. Evaluate the Integral: Depending on the values of k₁ and k₂, we have two cases. Case 1: k₁ ≠ k₂, $e^{k_2·t}B(t) = \frac{k_1A_0}{k_2-k_1}e^{(k_2-k_1)t} + C$. Case 2: k₁ = k₂, $e^{k_2·t}B(t) = k_1A_0t + C$
5. Solve for B(t).

For k₁ ≠ k₂, $B(t) = e^{-k_2·t}(\frac{k_1A_0}{k_2-k_1}e^{(k_2-k_1)t} + C)↭ B(t) = \frac{k_1A_0}{k_2-k_1}e^{-k_1t} + Ce^{-k_2·t}$

For k₁ = k₂ = k, $B(t) = e^{-k·t}(kA_0t + C)$

Interpretation of the Solution

1. 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.
2. Homogeneous Solution, $B_h(t) = Ce^{-k_2·t}$. Represents the natural decay of B independent of the substance A.

Initial Conditions

To determine C, we need an initial condition. If we assume that at time t = 0, the amount of B is B(0) = B₀, then:

• For k₁ ≠ k₂, $B(t) = B_0 = \frac{k_1A_0}{k_2-k_1}e^{-k_1·0} + Ce^{-k_2··0}↭ B_0 = \frac{k_1A_0}{k_2-k_1} + C ⇒[\text{Solve for C}] C = B_0 - \frac{k_1A_0}{k_2-k_1}$
• For k = k₁ = k₂, $B(0) = e^{-k·0}(kA_0·0 + C) ↭ B_0 = C$

Final Expression for B(t)

• For k₁ ≠ k₂, $B(t) = \frac{k_1A_0}{k_2-k_1}e^{-k_1t} + (B_0 - \frac{k_1A_0}{k_2-k_1})e^{-k_2·t}$
• For k = k₁ = k₂, $B(t) = e^{-k·t}(kA_0t + C) = e^{-k·t}(kA_0t + B_0)$

Physical Interpretation

• Initially, B increases as A decays into it. The term involving e^-k₁t represents this accumulation. Over time, B also decays, reducing its amount. The term involving e^-k₂t represents the natural decay of B.
• If k₁ > k₂, the accumulation of B is more pronounced (substance A decays rapidly) before it starts to decay (substance B decays more slowly). If k₁ < k₂, 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.

Special case $\frac{dy}{dt}-ay = q(t)$

This different equation is different from previous cases due to the negative coefficient of y. It is a first-order linear ODE and can exhibit exponentially growing solutions.

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

The constant C is determined by the initial condition y(t₀) = y₀. Small differences in y₀ can lead to large differences in y(t) as t increases due to the exponential growth.

Example: Constant Source Term q(t) = q₀

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

1. 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).
2. Simplify $y(t) = e^{at}(-\frac{q_0}{a}e^{-at} + C) = -\frac{q_0}{a} + Ce^{at}$
3. Apply initial condition. If y(0) = y₀, $y_0 = -\frac{q_0}{a} + Ce^{a·0} = -\frac{q_0}{a} + C [\text{Solve for C}] C = y_0 + \frac{q_0}{a}$.
4. Final solution: $y(t) = -\frac{q_0}{a} + (y_0 + \frac{q_0}{a})e^{at}$

if $y_0 + \frac{q_0}{a} ≠ 0$ and a > 0, the solution grows exponentially. The term $-\frac{q_0}{a}$ is constant. It represents the long-term behaviour (steady-state solution) if the exponential term diminishes (a < 0). The behavior heavily depends on the initial condition y₀. The term $y(t) = (y_0 + \frac{q_0}{a})e^{at}$ represent the transient behaviour of the solution.

As t→∞, the exponential term e^at dominates: $\lim_{t \to ∞} e^{at} = ∞$, leading the solution to diverge unless a < 0 or the initial condition is chosen such that $y_0 + \frac{q_0}{a} = 0$ (the exponential term vanishes and $y(t) = -\frac{q_0}{a}$, meaning the solution is constant for all t).

Solving First-Order Linear Ordinary Differential Equations with Sinusoidal Inputs

In this section, we will explore how to solve first-order linear ordinary differential equations (ODEs) with sinusoidal inputs using the method of complex exponentials and the integrating factor technique.

First-order linear ODEs with sinusoidal inputs frequently appear in engineering and physics, modeling systems like electrical circuits and mechanical vibrations. The general form of such an equation is: x ’ + kx = cos(wt), where:

• k is a positive constant representing the rate of decay (if k > 0) or growth (if k < 0).
• 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.

Step 1. 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]

Step 2. Formulating the Complex ODE. By substituting the complex exponential into the original differential equation, we obtain a complex ODE: z’ + kz = e^iwt analogous to the original equation. This equation is easier to solve, and we can later take the real part to find the solution to the original equation.

Step 3. 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:

1. Homogeneous Solution (Transient Solution): The solution that comes from the homogeneous equation, $ce^{-kt}$, it decays exponentially over time (if k > 0).
2. Particular Solution (Steady-State Solution). The solution that comes from the sinusoidal input is the particular solution, $\frac{1}{k+iw}e^{iwt}$.

Step 4: Extracting the Real Part to Find x(t). To find the real solution, we take the real part of z(t): x(t) = Re[z(t)] = $ce^{-kt}+ Re[\frac{1}{k+iw}e^{iwt}]$

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

x(t) = $ce^{-kt}+ Re[\frac{1}{k+iw}e^{iwt}] = ce^{-kt}+ Re[\frac{k-iw}{k^2+w^2}(cos(wt)+isin(wt))] = ce^{-kt}+ \frac{1}{k^2+w^2}[kcos(wt)+wsin(wt)]$ (we have just simplified -i²w = w)

Step 5: Analyzing the Solution

This expression consist of two parts: x(t) = x_h(t) + x_p(t) = $ce^{-kt}+ \frac{1}{k^2+w^2}[kcos(wt)+wsin(wt)]$ where C is determined by the initial conditions:

1. Homogeneous Solution (Transient Response): $x_h(t)=ce^{-kt}$, decays exponentially over time due to the e^-kt term if k > 0. It represents the natural response of the system without external input.
2. Particular Solution (Steady-State Response): $x_p(t)=\frac{1}{k^2+w^2}[kcos(wt)+wsin(wt)]$ oscillates with the same frequency w as the input. It represents the forced response due to the input cos(ωt).

Example: Solving x’ + kx = Fsin(wt)

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.

Step 1: Express sin(ωt) as a Complex Exponential. Using Euler’s formula: sin(wt) = Im(e^iwt).

Step 2. Formulate the complex ODE: z’ + kz = Fe^iwt.

Step 3: Solve the Complex ODE. Following a similar analysis, μ(t) = e^kt, $\frac{d}{dt}(e^{kt}z)=Fe^{(k+iw)t}, e^{kt}z = \int Fe^{(k+iw)t}dt + C ↭ e^{kt}z = \frac{F}{k+iw}e^{(k+iw)t} + C↭ e^{kt}z = \frac{F}{k+iw}e^{(k+iw)t} + C$, then (multiplying both sides by e^-kit) the general solution to the complex ODE is: $z = ce^{-kt}+ \frac{F}{k+iw}e^{iwt}$

Step 4: Extract the Imaginary Part: $\frac{F}{k+iw}e^{iwt} = \frac{F(k-iw)}{k^2+w^2}(cos(wt)+isin(wt))$

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

Solving x’ + kx = cos(wt) + 3sin(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 separately.

Split the equation into two separate equations: x’₁ + k₁ = cos(wt), x’₂ + kx₂ = 3sin(wt).

Find the particular solutions for each equation. From earlier analysis, 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 total particular solution is x_p = x_p1 + x_p2 = $\frac{1}{k^2+w^2}[kcos(wt)+wsin(wt)+3ksin(wt)-3wcos(wt)] = \frac{(k-3w)cos(wt)+(w+3k)sin(wt)}{k^2+w^2}$

Write the general solution (include the homogeneous solution x_h(t) = ce^-kt): x(t) = x_h(t) + x_p(t) = $ce^{-kt} + \frac{(k-3w)cos(wt)+(w+3k)sin(wt)}{k^2+w^2}$

Interpretation

• $x_h(t)=ce^{-kt}$ is the solution to the homogeneous equation x′ + kx = 0. It represents the transient response, it decays over time if k > 0.
• The particular solution x_p(t) = $\frac{(k-3w)cos(wt)+(w+3k)sin(wt)}{k^2+w^2}$ accounts for the input functions cos(wt) and 3sin(wt). It represents the steady-state response. It oscillates with frequency w.

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