The only way to learn mathematics is to do mathematics, Paul Halmos.
An algebraic equation is a mathematical statement that declares or asserts the equality of two algebraic expressions. These expressions are constructed using:
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., $\frac{dy}{dx} = 3x +5y$):
The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It states that if:
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 $p(x)=\frac{b(x)}{a(x)}\text{ and }q(x) = \frac{c(x)}{a(x)}$
A second-order linear homogeneous differential equation ODE with constant coefficients is a differential equation of the form: y'' + Ay' + By = 0 where:
This equation is homogeneous, meaning that there are no external forcing terms (like a function of t) on the right-hand side.
To explain how the equation y’ + ky = q(t) refers to the theorem on solving inhomogeneous linear differential equations, let’s break down the solution process and compare it to the concepts of complementary and particular solutions.
We start with the first-order linear inhomogeneous differential equation: y’ + ky = q(t) where:
The goal is to solve this differential equation and understand how it relates to the theorem on solving inhomogeneous linear differential equations, specifically in terms of finding the complementary and particular solutions.
Step 1: Solving the Homogeneous Equation
The homogeneous equation corresponding to the given differential equation is:: 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’’. Thus, the complementary solution is: y_{c} = Ce^{-kt}.
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} = $e^{-kt}\int e^{kt}q(t)dt$
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(t) = y_{p} + y_{c} = $e^{-kt}\int e^{kt}q(t)dt + Ce^{-kt}$
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:
The total solution is a sum of the particular and complementary solutions, just as the theorem suggests.
Behaviour as t → ∞. When k > 0, the term $Ce^{-kt}$, representing the complementary solution (or transient part), decays to zero as t → ∞. 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 disappear.
On the other hand, when k < 0, the term $Ce^{-kt}$ in the complementary solution grows exponentially as t → ∞ (since -kt becomes positive), causing the solution y(t) to grow without bond and making the above analysis irrelevant as the transient part dominates and the particular solution y_{p}(t) becomes negligible. Physically, the system exhibits unbounded growth or instability.
1: Solve the Homogeneous Equation. Write the Homogeneous Equation: y’ + 2y = 0.
Separate Variables: $\frac{dy}{dt} = -2y ⇒ \frac{dy}{y} = -2dt$
Integrate both sides: $\int \frac{dy}{y} = \int-2dt ⇒ ln|y| = -2t + C'$
Solve for y: $|y| = e^{-2t+C’} = C’’e^{-2t}$ where C’’ = e^{C’}. $y = ± C’’e^{-2t} = Ce^{-2t}$ where C = ±C’'.
Complementary Solution: $y_c(t) = Ce^{-2t}$
Step 2: Find a Particular Solution
Compute the Integrating Factor: $μ(t) = e^{\int 2dt} = e^{2t}$
Multiply Both Sides: $e^{2t}y’ + 2e^{2t}y = e^{2t}e^{3t}$
Simplify Left-Hand Side: $\frac{d}{dx}(e^{2t}y) = e^{5t}$
Integrate both sides: $e^{2t}y = \int e^{5t}dt + C ↭ e^{2t}y = \frac{1}{5}e^{5t} + C$
Solve for y_{p}(t). Ignoring the constant C (since it’s part of 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) = $Ce^{-2t} + \frac{1}{5}e^{3t}$
1: Solve the Homogeneous Equation. Write the Homogeneous Equation: y’ + 2ty = 0.
Separate Variables: $\frac{dy}{dt} = -2ty ⇒ \frac{dy}{y} = -2tdt$
Integrate both sides: $\int \frac{dy}{y} = \int-2tdt ⇒ ln|y| = -t^2 + C'$
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’'.
Complementary Solution: $y_c(t) = Ce^{-t^2}$
Step 2: Find a Particular Solution
Compute the Integrating Factor: $μ(t) = e^{\int 2tdt} = e^{t^2}$
Multiply Both Sides: $e^{-t^2}y’ + 2e^{-t^2}y = e^{t^2}4t$
Simplify Left-Hand Side: $\frac{d}{dt}(e^{t^2}) = e^{t^2}4t$
Integrate both sides: $e^{2t}y = \int 4te^{t^2}dt + C ↭ e^{t^2}y = 2e^{t^2} + C$
Let u = t^{2} ⇒ du = 2tdt, $tdt = \frac{du}{2} \int 4te^{t^2}dt = \int 4·\frac{du}{2}·e^u = \int 2e^udu = 2e^u + C = 2e^{t^2} + C$
Solve for y_{p}(t). Ignoring the constant C (since it’s part of the complementary solution). Subtract $2e^{t^2}$ from both sides $(e^{t^2}y -2e^{t^2}) = C ↭ e^{t^2}(y-2) = C ↭[\text{Divide both sides by } e^{t^2}] y - 2 = C·e^{-t^2}$. Since $C·e^{-t^2}$ is part of the complementary solution, we can set and find 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.
Analysis of the Solution
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:
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.
The general solution to the inhomogeneous differential equation consists of two parts:
Thus, the general solution is: y = y_{p} + y_{c} = y_{p} + c_{1}y_{1} + c_{2}y_{2} where:
To find the complementary solution y_{c}(t), we solve the homogeneous equation: y′′ +Ay’ +By = 0.
We assume solutions of the form y(t) = e^{rt}, where r is a constant. Substituting into the homogeneous equation gives the characteristic equation: $r^2e^{rt}+ Are^{rt} + Be^{rt} = 0↭ [\text{Dividing by } e^{rt}] r^2 + Ar + B = 0$.
The roots of this quadratic equation, denoted as r_{1}, r_{2}, 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:
Where:
Characteristic roots | Solutions | Stability Condition |
---|---|---|
r₁ ≠ r₂ | ||
(distinct real roots) | C₁e^{r₁t}+C₂e^{r₂t} | r₁< 0, r₂< 0 |
r = r₁ = r₂ (repeated real root) | (C₁+C₂t)e^{rt} | r< 0 |
r = a ± bi (complex roots) | e^{at}(C₁cos(bt)+C₂sin(bt)) | a < 0 |
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 c_{1}y_{1} + c_{2}y_{2} goes to zero as t goes to infinity (t→ ∞) ∀c_{1}, c_{2} (for any values of c_{1} and c_{2})?
Distinct real roots r_{1} and r_{2}. The complementary solution for distinct real roots is: $y_c(t) = C_1e^{r_1t}+C_2e^{r_2t}$.
For the solution to decay to zero as t→∞, both exponential terms must tend to zero. This occurs if both r_{1} and r_{2} are negative. Therefore, the ODE is stable if both characteristic roots are negative.
If either root is positive, the corresponding exponential term will grow exponentially, leading to an unstable solution.
Repeated real root r_{1} = r_{2} = r. When the characteristic equation has a repeated real root r_{1} = r_{2} = 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 and the term t·e^{rt} also approaches zero.
If r = 0, the solution becomes y_{c}(t) = C_{1} + C_{2}t, which does not decay to zero unless C_{1} = C_{2} = 0, so it’s unstable.
If r > 0, the exponential term grows, leading to an unstable solution
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 behavior of the solution depends on the real part a. If a < 0, the exponential factor e^{at} causes the oscillations to decay to zero 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, leading to a marginally stable or unstable system.
If α > 0, the exponential term grows, making the solution unstable.
To summarize, the second-order y'' + Ay' + By = f(t) is stable if the characteristic roots have negative real parts.
In physical systems, stability often reflects how a system responds to disturbances or external forces:
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: mr^{2} + br + k = 0.
Stability analysis:
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.
Characteristic Equation: $Lr^2 + Rr + \frac{1}{C} = 0$
Stability Analysis