First-order linear inhomogeneous differential equation

The only way to learn mathematics is to do mathematics, Paul Halmos.

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.

Second-order Linear Homogeneous ODE’s with Constant Coefficients

A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0, where A and B are constants.

To solve this ODE, we seek two linearly independent solutions y₁(t) and y₂(t). The general solution is then a linear combination of these solutions: $c_1y_1 + c_2y_2$ where c₁ and c₂ are two arbitrary constants determined by initial conditions. The key to solving the ODE is the characteristic equation, whose roots determine the behavior of the solutions.

First-order linear inhomogeneous differential equation

A first-order linear inhomogeneous differential equation is an equation of the form: y’ + ky = q(t). Our goal is to solve this differential equation and understand how it relates to the general theorem on solving inhomogeneous linear differential equations, particularly in terms of finding the complementary and particular solutions.

We start with the first-order linear inhomogeneous differential equation: y’ + ky = q(t) where:

• y is the unknown function of the variable t.
• y′ is the derivative of y with respect to t.
• k is a constant coefficient.
• q(t) is a given function of t, known as the inhomogeneous term, forcing term, or input.

Step 1: Solving the Homogeneous Equation

The homogeneous equation corresponding to the given differential equation is obtained by setting the inhomogeneous term q(t) = 0: y’ + ky = 0. This equation describes the system’s behavior without any external forcing.

Solving the Homogeneous Equation. y’ + ky = 0 ↭ $\frac{dy}{dt} = -ky$ ⇒[To solve this, we separate variables:] $\frac{dy}{y}=-kdt$⇒[Integrating both sides:] $\int \frac{dy}{y} = \int -kdt$ ↭ ln|y| = -kt + C’ where C’ is the constant of integration.

Exponentiating both sides: |y| = e^C’e^-kt = C’’e^-kt where C’’ = e^C’ ↭ y = ±C’’e^-kt = Ce^-kt where C = ± C’’ (which can be any nonzero real constant). Thus, the complementary solution is: y_c = Ce^-kt and C is an arbitrary constant determined by initial conditions.

Step 2: Solving the Inhomogeneous Equation

To find a particular solution y_p to the inhomogeneous equation: y’ + ky = q(t), we can use the method of integrating factors.

We multiply both sides of the equation by the integrating factor μ(t) = $e^{\int kdt} = e^{kt}$: e^kt·y’ + e^kt·ky = e^kt·q(t)

This simplifies the left-hand side to: $\frac{d}{dt}(e^{kt}·y)=e^{kt}q(t)$.

Now, integrate both sides with respect to t: $e^{kt}·y = \int e^{kt}q(t)dt$.

Thus, the particular solution is: y_p(t) = $e^{-kt}\int e^{kt}q(t)dt$

Step 3: Writing the General Solution

The general solution to the linear inhomogeneous differential equation of the form is the sum of the complementary and particular solution: y = y_p + y_c where:

• y_p is a particular solution, any specific solution to the inhomogeneous equation y’ + ky = q(t). It represents the system’s response to the external forcing term q(t). It does not contain arbitrary constants (any constant of integration is absorbed into the complementary solution).
• y_c is the complementary solution. It solves the corresponding homogeneous equation y’ + ky = 0. It represents the homogeneous behavior of the system, in other words, how the system would behave in the absence of the external forcing (q(t) = 0).

y(t) = y_p + y_c = $e^{-kt}\int e^{kt}q(t)dt + Ce^{-kt}$

Connection to the General Theorem on Inhomogeneous Linear Differential Equations

This aligns with the general theorem on solving linear inhomogeneous differential equations, which states that:

According to the theorem on linear inhomogeneous equations, the general solution consists of:

• A particular solution y_p, which in this case is $e^{-kt}\int e^{kt}q(t)dt$. Any specific solution to the inhomogeneous equation y′ +ky = q(t). It represents the system's response to the external forcing term q(t) and not contain arbitrary constants (any constant of integration is absorbed into the complementary solution).
• A complementary solution y_c. It solves the homogeneous equation y’ + ky = 0, which in this case is $Ce^{-kt}$. It represents the homogeneous behavior of the system. It describes how the system would behave in the absence of external forcing (q(t) = 0). Contains an arbitrary constant C determined by initial conditions.

The total solution is a sum of the particular and complementary solutions, just as the theorem suggests. y(t) = y_p + y_c = $e^{-kt}\int e^{kt}q(t)dt + Ce^{-kt}$

Behaviour as t → ∞. Understanding the long-term behavior of the solution helps in interpreting the physical implications.

Case 1. When k > 0, the term $Ce^{-kt}$, representing the complementary solution (or transient part), decays to zero as t → ∞, $\lim_{t \to ∞}y_c(t) = \lim_{t \to ∞}Ce^{-kt} = 0$. The solution y(t) approaches the particular solution y_p(t), representing a steady-state, which depends on q(t). This reflects the physical idea that over time, the system stabilizes and the transient effects (due to initial conditions) disappear or diminish.

Case 2. When k < 0, the term $Ce^{-kt}$ in the complementary solution grows exponentially as t → ∞ (since -kt becomes positive) $\lim_{t \to ∞}y_c(t) = \lim_{t \to ∞}Ce^{-kt} = ∞$ (if C > 0), causing the solution y(t) to grow without bound and making the above analysis irrelevant as the exponentially growing complementary solution (transient part) dominates, making the particular solution y_p(t) negligible in comparison. Physically, the system exhibits unbounded growth or instability. The external forcing q(t) has little effect on the long-term behavior.

Solved examples

• y’ + 2y = e^3t.

Step 1: Solve the Homogeneous Equation. The homogeneous equation is obtained by setting the right-hand side to zero: y’ + 2y = 0.

1. Separate Variables: $\frac{dy}{dt} = -2y ⇒ \frac{dy}{y} = -2dt$

2. Integrate both sides: $\int \frac{dy}{y} = \int-2dt ⇒ ln|y| = -2t + C'$

3. Solve for y: $|y| = e^{-2t+C’} = C’’e^{-2t}$ where C’’ = e^C’. $y = ± C’’e^{-2t} = Ce^{-2t}$ where C = ±C’’(an arbitrary constant).

4. Complementary Solution: $y_c(t) = Ce^{-2t}$

Step 2: Find a Particular Solution

We will use the method of integrating factors to find a particular solution y_p(t).

1. Compute the Integrating Factor: $μ(t) = e^{\int 2dt} = e^{2t}$

2. Multiply Both Sides: $e^{2t}y’ + 2e^{2t}y = e^{2t}e^{3t}$

3. Simplify Left-Hand Side: $\frac{d}{dt}(e^{2t}y) = e^{5t}$

4. Integrate both sides: $e^{2t}y = \int e^{5t}dt + C ↭ e^{2t}y = \frac{1}{5}e^{5t} + C$

5. Solve for y_p(t). Ignoring the constant C (it will be accounted for in the complementary solution): y_p(t) = $e^{-2t}\frac{1}{5}e^{5t} = \frac{1}{5}e^{3t}$

Step 3: Write the General Solution

y(t) = y_c(t) + y_p(t) = $Ce^{-2t} + \frac{1}{5}e^{3t}$

Step 4: Analysis of the solution

1. Complementary solution. Since e^-2t decays exponentially to zero as t→∞, the complementary solution approaches zero. This term represents the transient behavior of the system.
2. Particular solution. Since e^3t grows exponentially as t→∞, the particular solution increases without bound. This term represents the steady-state response of the system to the forcing term e^3t.
3. For larger t, the particular solution dominates the behavior of y(t). Therefore, y(t) ≈ $\frac{1}{5}e^{3t}$ as t → ∞.

• y’ + 2ty = 4t

Step 1: Solve the Homogeneous Equation. Write the Homogeneous Equation: y’ + 2ty = 0.

2. Separate Variables: $\frac{dy}{dt} = -2ty ⇒ \frac{dy}{y} = -2tdt$

3. Integrate both sides: $\int \frac{dy}{y} = \int-2tdt ⇒ ln|y| = -t^2 + C'$

4. Solve for y: $|y| = e^{-t^2+C’} = C’’e^{-t^2}$ where C’’ = e^C’. $y = ± C’’e^{-t^2} = Ce^{-t^2}$ where C = ±C’'.

5. Complementary Solution: $y_c(t) = Ce^{-t^2}$

Step 2: Find a Particular Solution. We will use the method of integrating factors.

1. Compute the Integrating Factor: $μ(t) = e^{\int 2tdt} = e^{t^2}$

2. Multiply Both Sides: $e^{-t^2}y’ + 2e^{-t^2}y = e^{t^2}4t$

3. Simplify Left-Hand Side: $\frac{d}{dt}(e^{t^2}·y) = e^{t^2}4t$

4. Integrate both sides: $e^{2t}y = \int 4te^{t^2}dt + C ↭ e^{t^2}y = 2e^{t^2} + C$

   Let u = t² ⇒ du = 2tdt, $tdt = \frac{du}{2}, \int 4te^{t^2}dt = \int 4·\frac{du}{2}·e^u = \int 2e^udu = 2e^u + C = 2e^{t^2} + C$

5. Solve for y_p(t). Ignoring the constant C (it will be accounted for in the complementary solution). Subtract $2e^{t^2}$ from both sides $(e^{t^2}y -2e^{t^2}) = C ↭ e^{t^2}(y-2) = C ↭[\text{Divide both sides by } e^{t^2}] y - 2 = C·e^{-t^2}↭ y = 2 + C·e^{-t^2}$. Since $C·e^{-t^2}$ is already part of the complementary solution, we can set C = 0 and take as the particular solution: y_p(t) = 2.

Step 3: Write the General Solution

y(t) = y_c(t) + y_p(t) = $C·e^{-t^2} + 2$ where C is a constant determined by initial or boundary conditions.

Step 4. Analysis of the Solution

1. Complementary Solution ($C·e^{-t^2}$) decays to zero as t→∞ because e^-t2 decreases exponentially (much faster than any exponential function).
2. Particular Solution. The particular solution is a constant. It represents the steady-state response to the forcing term 4t.
3. Overall Behavior: For large t, y(t) approaches the constant value 2. Therefore, y(t) ≈ 2 as t → ∞.

Stability Criteria for Second-Order Linear Differential Equations with Constant Coefficients

We are exploring the stability of solutions to second-order linear differential equation with constant coefficients of the form: y’’ + Ay’ + By = f(t) where:

• A and B are constant.
• y(t) is the unknown function of time.
• y′ and y′′ are the first and second derivatives of y with respect to t, respectively
• f(t) is the forcing term, representing an external input or driving force.

Understanding the stability of such equations is crucial in various fields like mechanical engineering, electrical engineering, and physics, as it helps predict the long-term behavior of dynamic systems, determining whether they will settle into a steady state or exhibit unbounded growth or sustained oscillations.

General Solution of the Differential Equation

The general solution to the inhomogeneous differential equation consists of two parts:

1. Particular Solution y_p. A specific solution that solves the inhomogeneous equation y’’ + Ay’ + By = f(t). This solution accounts for the system’s response to the external forcing term f(t). It does not contain arbitrary constants.
2. Complementary solution y_c (or homogeneous solution). The general solution to the corresponding or associated homogeneous equation y’’ + Ay’ + By = 0. This solution represents the system’s natural behavior without external input. It contains arbitrary constants determined by initial conditions.

Therefore, the general solution is: y = y_p + y_c = y_p + c₁y₁ + c₂y₂ where:

• y_p(t) addresses the external forcing term f(t).
• y_c = c₁y₁ + c₂y₂ are solutions to the homogeneous equation. It accounts for the natural behavior of the system without external input.
• c₁ and c₂ are constants determined by the initial conditions.

Complementary Solution and Stability

To analyze stability, we focus on the complementary solution y_c(t) because it determines how the system responds over time without external forcing.

Solving the Homogeneous Equation

To find the complementary solution y_c(t), we solve the homogeneous equation: y′′ +Ay’ +By = 0.

Assuming a solution of the form y(t) = e^rt, where r is a constant, we substitute into the homogeneous equation and gives us the characteristic equation: $r^2e^{rt}+ Are^{rt} + Be^{rt} = 0↭ [\text{Dividing by } e^{rt} ≠ 0] r^2 + Ar + B = 0$.

Form of the Complementary Solution

The roots of this quadratic equation, denoted as r₁, r₂, or simply r, determine the form of the complementary solution. The behavior of the complementary solution and thus, the stability of the ODE depends on these roots.

Let’s examine the different types of characteristic roots and their corresponding solutions:

1. Distinct Real Roots (r₁ ≠ r₂): y_c(t) = c₁e^r₁t + c₂e^r₂t. The solution is stable if both r₁ < 0, r₂ < 0.
2. Repeated Real Roots (r₁ = r₂ = r): y_c(t) = (c₁ + c₂t)e^rt. The solution is stable if r < 0.
3. Complex Roots (r = α + βi): y_c(t) = e^αt(C₁cos(βt) + C₂sin(βt)). The solution is stable if α = Re(r) < 0.

Where:

• C₁ and C₂ are constants determined by initial conditions.
• α = Re(r) is the real part of the complex root.
• β = Im(r) is the imaginary part of the complex root.

In this context, stability refers to the behavior of the complementary solution y_c(t) as time t approaches infinity (t→ ∞). Specifically, we are interested in the question: When does the term c₁y₁ + c₂y₂ goes to zero as t goes to infinity (t→ ∞) ∀c₁, c₂ (for any values of c₁ and c₂)?↭$\lim_{t \to ∞}y_c(t) = 0~ ∀c_1,c_2$.

• Stable Solution. If y_c(t) = c₁y₁ + c₂y₂ → 0 as t→ ∞ for all initial conditions (i.e., for all values of c₁ and c₂), the solution is said to be stable. The transient part of the solution, y_c(t) decays over time, the system settle into a steady-state behaviour described by the particular solution y_p. External disturbances or initial conditions have diminishing effects as time progresses.
• Unstable Solution. If y_c(t) does not approach zero as t → ∞, meaning it grows without bound or oscillates indefinitely, the solution is unstable. The transient part of the solution persists or grows over time. The system does not settle into a steady state. Initial conditions or disturbances have lasting or increasing effects.

Analysis of Stability Based on Characteristic Roots

1. Distinct real roots r₁ and r₂. The complementary solution for distinct real roots is: $y_c(t) = C_1e^{r_1t}+C_2e^{r_2t}$.

   For the solution to decay exponentially to zero as t→∞, both exponential terms must tend to zero. This occurs if both roots r₁ and r₂ are negative. Therefore, the ODE is stable if both characteristic roots are negative.

   If either root r₁ > 0 or r₂ > 0 is positive, the corresponding exponential term will grow exponentially, leading to an unstable solution.

2. Repeated real root r₁ = r₂ = r. When the characteristic equation has a repeated real root r₁ = r₂ = r, the complementary solution takes the form: $y_c = (C_1+C_2t)e^{rt}$.

   In this case, the solution is stable if r < 0 because e^rt→ 0 (decays exponentially) and the term t·e^rt also approaches zero. The polynomial term (c₁ + c₂t) may grow with t, but since it’s multiplied by a decaying exponential, the overall solution still approaches zero.

   If r = 0, the solution becomes y_c(t) = C₁ + C₂t, which does not decay to zero unless C₁ = C₂ = 0, so it’s unstable.

   If r > 0, the exponential term grows exponentially, leading to instability.

3. Complex Roots r = α± βi. When the characteristic roots are complex, the complementary solution is of the form: $y_c(t) = e^{at}(C_1cos(bt)+C_2sin(bt))$. Here, the oscillatory terms cos(bt) and sin(bt) are multiplied by the exponential factor e^at. The exponential term dictates the amplitude of the oscillations.

   The behavior of the solution depends on the real part a. If a < 0, the exponential term e^at decays, causing the oscillations to diminish over time, and the solution is stable.

   The oscillatory terms cos(βt) and sin(βt) are bounded between −1 and 1, so the exponential term dictates the overall behavior.

   If α = 0, the solution oscillates indefinitely without decaying, the amplitude remains constant, leading to sustained oscillations (this is a case of marginal stability).

   If α > 0, the exponential term grows, the oscillations increase in amplitude, making the solution unstable.

To summarize, the second-order y'' + Ay' + By = f(t) is stable if the characteristic roots have negative real parts.

Physical Interpretation of Stability

In physical systems, understanding the concept of stability is crucial for predicting how physical systems behave over time. It refers to the system’s response to initial disturbances or external forces and whether the system returns to equilibrium or diverges as time progresses.

Physical systems often exhibit two primary behaviors in response to stimuli:

• Transient behavior (given by y_c) represents how the system responds initially to initial conditions or disturbances. If the system is stable, this transient behavior will decay over time, allowing the system to settle. Dominates the solution in the short term but diminishes as t → ∞.
• Steady-state behavior (given by y_p) represents the long-term behavior of the system after the transient effects have subsided and becomes prominent after transient effects have decayed. Determines how the system responds to continuous external inputs over an extended period.

Examples

• Spring-mass-dashpot system.

Stability means that any initial oscillations or displacements due to an external force or displacement will eventually die out, allowing the system to return to equilibrium. This happens when the damping force, represented by the dashpot (b), is sufficient to counteract the effects of any initial energy input, such as an external force or displacement.

An equation modeling a damped harmonic oscillator (e.g., a mass attached to a spring and dashpot) is: my’’ + by’ + ky = F(t) where: m: Mass; b: Damping coefficient (dashpot); k: Spring constant; F(t): External force.

Characteristic equation: mr² + br + k = 0. The roots of the characteristic equation, r, determine the system’s behavior. We analyze the discriminant D = b² -4mk.

Stability analysis:

1. Overdamped (b² > 4mk, D > 0): Two distinct real negative (if b > 0) roots r₁ and r₂. Complementary solution: $y_c(t) = c_1e^{r_1t}+c_2e^{r_2t}$. Exponential decay without oscillations. The system returns to equilibrium slowly, hence the solution is stable.
2. Critically Damped (b² = 4mk, D = 0): Repeated real negative root $r=\frac{-b}{2m}$. Complementary solution: $y_c(t)=(c_1+c_2t)e^{rt}$. It is the the fastest to return to equilibrium without oscillations. The system decays to zero as quickly as possible, hence the solution is stable.
3. Underdamped (b² < 4mk, D < 0): Complex roots (α + βi) with negative real part, $α = \frac{-b}{2m}<0, β = \frac{\sqrt{4mk-b^2}}{2m}$. Complementary solution: $y_c(t)=e^{αt}(c_1cos(βt)+c_2sin(βt))$. It describes an oscillatory motion with decreasing amplitude. The system oscillates while gradually returning to equilibrium, hence solution is stable with oscillations.
4. No Damping (b = 0, D = -4mk < 0): Purely imaginary roots r = ±iw where $w=\frac{k}{m}$. Complementary solution: $y_c(t)=(c_1cos(wt)+c_2sin(wt))$. There are sustained oscillations with constant amplitude. The system continues to oscillate indefinitely. It is marginally stable, the system remains bounded but does not return to equilibrium.

• Electrical RLC Circuit

In electrical circuits, stability implies that any fluctuations in voltage or current will eventually stabilize, and the circuit will reach a steady operating state. The stability of an electrical circuit is influenced by factors such as the nature of the circuit (e.g., resistive, inductive, or capacitive), the level of damping (e.g., due to resistors, inductors, or capacitors), and the operating conditions.

An RLC circuit is governed by: $Lq’’ + Rq’ + \frac{1}{C}q = E(t)$ where L: Inductance; R: Resistance; C: Capacitance; q(t): Charge; E(t): External voltage source.

Consider the homogeneous equation (E(t) = 0): $Lq’’ + Rq’ + \frac{1}{C}q = 0$. Assuming a solution q(t) = e^rt, we substitute into the homogeneous equation: $Lq(r^2e^{rt}) + R(re^{rt}) + \frac{1}{C}e^{rt} = 0$. Divide both sides by e^rt.Characteristic Equation: $Lr^2 + Rr + \frac{1}{C} = 0$. Its roots determine the behaviour of the system, we analyze the discriminant: $D =R^2 -\frac{4L}{C}$:

Stability Analysis

1. Overdamped, $R^2 > 4\frac{L}{C}$ (D > 0). Two distinct real negative (if R > 0) roots; solution is stable. Complementary solution: $q_c(t) = c_1e^{r_1t}+c_2e^{r_2t}$. There’s an exponential decay of current and charge without oscillations. The circuit returns to steady state slowly, hence the system is stable.
2. Critically Damped, $R^2 = 4\frac{L}{C}$ (D = 0). Repeated real negative root $r=\frac{-R}{2L}$. Complementary solution: $q_c(t)=(c_1+c_2t)e^{rt}$. This system shows the fastest return to steady state without oscillations, hence the solution is stable.
3. Underdamped, $R^2 < 4\frac{L}{C}$ (D < 0). Complex roots (α + βi) with negative real part, $α = \frac{-R}{2L}<0, β = \frac{\sqrt{4\frac{L}{C}-R^2}}{2L}$. Complementary solution: $q_c(t)=e^{αt}(c_1cos(βt)+c_2sin(βt))$. There’s an oscillatory motion with decreasing amplitude. The system oscillates while gradually returning to equilibrium, hence the solution is stable with oscillations.
4. No Resistance (R = 0), $-4\frac{L}{C} < 0$. Purely imaginary roots (r = ±iw) where w = $\sqrt{\frac{1}{LC}}$. Complementary solution: $q_c(t)=c_1cos(βt)+c_2sin(βt)$. There are sustained oscillations of current and voltage with constant amplitude. The circuit continues to oscillate indefinitely (marginally stable).

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