The essence of mathematics is not to make simple things complicated, but to make complicated things simple, S. Gudder
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.
An inhomogeneous second-order linear differential equation is a differential equation of the form: y'' + p(x)y' + q(x)y = f(x) where:
The goal is to find y(x), known as the response or output, that satisfies this equation.
To solve the inhomogeneous equation, we proceed in two main steps:
Solve the Associated Homogeneous Equation. Before addressing the inhomogeneous equation, we first solve the corresponding homogeneous equation: y’’ + p(x)y’ + q(x)y = 0.
This is called the associated or reduced homogeneous equation. Its solutions are denoted as y_{c}(x) = c_{1}y_{1}(x) + c_{2}y_{2}(x), where: y_{1} and y_{2} are linearly independent solutions of the homogeneous equation. These solutions are known as the complementary solutions. c_{1} and c_{2} are arbitrary constants determined by initial or boundary conditions.
c_{1} and c_{2} are arbitrary constants determined by initial or boundary conditions.
Find a Particular Solution. Next, we find a particular solution y_{p}(x) to the inhomogeneous equation: y’’ + p(x)y’ + q(x)y = f(x). The particular solution y_{p}(x) satisfies the inhomogeneous equation and accounts for the presence of the forcing term f(x).
The general solution to the inhomogeneous equation is the sum of the complementary solution y_{c} and a particular solution y_{p}: y(x) = y_{p}(x) + y_{c}(x) = y_{p}(x) + c_{1}y_{1}(x) + c_{2}y_{2}(x).
The spring-mass-dashpot system. The equation describing the motion of a damped mass-spring system with external forcing is: mx’’+bx’ + kx = f(t) where:
m: Mass of the object.
b: Damping coefficient (related to the dashpot).
k: Spring constant.
x(t): Displacement of the mass at time t.
f(t): External force applied to the system. The system is said to be forced or influenced by an external force when f(t) ≠ 0, and passive when f(t) = 0 (the system evolves solely based on its initial conditions and inherent properties).
Simple Electrical Circuit (RLC Circuit). An RLC circuit consists of a resistor (R), an inductor (L), a capacitor (C), and a voltage source ε(t). According to Kirchhoff’s Voltage Law (KVL), the sum of voltage drops around a closed loop equals the applied voltage: $L\frac{di}{dt}+Ri+\frac{q}{C} = ε(t)$ where i(t) is the current flowing through the circuit and q(t) is the charge of the capacitor, with $\frac{dq}{dt} = i(t)$, meaning the rate of change of charge (dq/dt) with respect to time is equal to the current (i).
Voltage Drops:
Therefore, $\frac{dq}{dt} = i$ and applying KVL to the RLC circuit gives us the equation: $L\frac{di}{dt}+Ri+\frac{q}{C} = ε(t)$. Differentiating this equation guves: $L\frac{d^2i}{dt^2}+R\frac{di}{dt} + \frac{i}{C} = ε’(t)$ ↭[Notation] Li’’ + Ri’ + ^{i}⁄_{C} = ε’(t). This equation governs the flow of current in the circuit in response to an applied voltage ε(t).
Let L = D^{2} +pD + q =[Notation] $\frac{d^2}{dx^2}+p(x)\frac{d}{dx} + q(x)$ be the differential operator and let our inhomogeneous second-order linear differential equation be y’’ + p(x)y’ + q(x)y = f(x).
Then, the inhomogeneous differential equation becomes: Ly = f(x). And the associated homogeneous equation is: Ly = 0.
The general solution to the inhomogeneous equation is: y(x) = y_{p} + y_{c} = y_{p} + c_{1}y_{1} + c_{2}y_{2} where:
The reason for adding the complementary solution is that the inhomogeneous equation consists of two parts:
The complementary solution y_{c} accounts for the behavior of the system without the forcing term, while the particular solution y_{c} responds to the input or forcing term f(x).
Proof.
Let’s prove why the general solution is of the form y(x) = y_{p} + c_{1}y_{1} + c_{2}y_{2}
Step 1: Proving the form is a solution
We need to prove that y = y_{p} + y_{c} = y_{p} + c_{1}y_{1} + c_{2}y_{2} is a solution to the original equation y’’ + p(x)y’ + q(x)y = f(x) ↭ Ly = f(x)
L(y) = L(y_{p} + c_{1}y_{1} + c_{2}y_{2}) =[L is a linear operator] L(y_{p}) + L(c_{1}y_{1} + c_{2}y_{2}) =[y_{c} = c_{1}y_{1} + c_{2}y_{2} is a solution to the associated homogeneous equation, i.e., L(c_{c}) = L(c_{1}y_{1} + c_{2}y_{2}) = 0. We knot that L(y_{p}) = f(x) because y_{p} is a particular solution ] f(x) + 0 = f(x)
Step 2: Proving no other solutions exist
Second, we need to show that there are no other solutions. Now, assume that u(x) is another solution to the inhomogeneous equation ⇒[By definition, u is a solution and y_{p} is a particular solution] L(u) = f(x), L(y_{p}) = f(x).
Subtract these two equations: L(u)-L(y_{p}) = f(x) -f(x) = 0.
Since L is a linear operator, this simplifies to: L(u-y_{p}) = 0. This means that u-y_{p} is a solution to the associated homogeneous solution Ly = 0.
Therefore, u-y_{p} must be a linear combination of y_{1} and y_{2}: $∃\tilde{c_1},\tilde{c_2}: u-y_p = \tilde{c_1}y_1 + \tilde{c_2}y_2 ⇒u = y_p + \tilde{c_1}y_1 + \tilde{c_2}y_2$, which is exactly of the form y_{p} + c_{1}y_{1} + c_{2}y_{2}.
This proves that the general solution y = y_{p} + c_{1}y_{1} + c_{2}y_{2} includes all possible solutions to the inhomogeneous equation.
Assumed Form: y_{p}(x) = $xe^{2x}$. Compute Derivatives: y_{p}’(x) = $e^{2x} + 2xe^{2x}$, y_{p}’’(x) = $2e^{2x} + 2e^{2x} + 2xe^{2x}·2 = 4e^{2x}+4xe^{2x}$
Substitute into the Inhomogeneous Equation: $[4e^{2x}+4xe^{2x}] -3[e^{2x} + 2xe^{2x}] +2xe^{2x} = e^{2x} ↭ (4-3)e^{2x} + (4-6+2)xe^{2x} = e^{2x}$. Therefore, the particular solution is $y_p(x) = xe^{2x}$, and the general solution is $y(x) = y_c(x) + y_p(x) = c_1e^x + c_2e^{2x} + xe^{2x}$
Compute derivatives: $y_p’(t) = 2Ae^{2t}, y_p’’(t) = 4Ae^{2t}$.
Substitute into the Inhomogeneous Equation: $4Ae^{2t} -4Ae^{2t} -3Ae^{2t} = 3e^{2t} ↭[\text{Simplify}] -Ae^{2t} = e^{2t}⇒[\text{Divide both sides by }e^{2t}] A = -1$. Therefore, the particular solution is $y_p = -e^{2t}$, and the general solution is $y(x) = y_c(x) + y_p(x) = c_1e^{3t} + c_2e^{-t} -e^{2t}$
When choosing the form of the particular solution, consider the nature of the non-homogeneous term. If the non-homogeneous term is of the form e^{rt} and r is not a root of the characteristic equation, we can assume y_{p} = Ae^{rt}. If r is a root of the characteristic equation, we must multiply by t enough times to make y_{p}, linearly independent from the complementary solution.
Table of Common Non-Homogeneous Terms and Corresponding Particular Solution Forms
Non-homogeneity | Guess |
---|---|
$e^{rt}$ | $Ae^{rt}$ |
sin(rt) or cos(rt) | Asin(rt) + Bcos(rt) |
Degree n polynomial | A_{0} + A_{1}t + ··· + A_{n}t^{n} |
Multiply the assumed form by t^{s} as needed to avoid duplication with terms in the complementary solution.
Compute derivatives: y_{p}’(x) = 2Ax + B, y_{p}’’(x) = 2A.
Substitute into the Inhomogeneous Equation: 2A + 5(2Ax + B) + 6(Ax^{2} + Bx + C) = x^{2} ↭ 2A + 10Ax + 5B + 6Ax^{2} + 6Bx + 6C = x^{2} ↭[Group like terms:] (6A)x^{2} + (10A +6B)x + (2A + 5B + 6C) = x^{2}
Set up a system of equations by equating coefficients:
$\begin{cases} 6A = 1~ (i) \\ 10A +6B = 0 ~ (ii)\\ 2A + 5B + 6C~ (iii)\end{cases}$
Solve for A, B, and C:
(i) ⇒ A = ^{1}⁄_{6}. (ii) $10·\frac{1}{6} + 6B = 0↭[\text{Multiply by 6}] 10 + 36B = 0 ⇒ B = \frac{-10}{36} = \frac{-5}{18}$. (iii) $2·\frac{1}{6}+5\frac{-5}{18} +6C = 0 ↭[\text{Multiply by 18}] 6 -25 + 108C = 0↭ -19 +108C = 0 ⇒ C = \frac{19}{108}$. Therefore, the particular solution is $y_p = Ax^2+Bx+C = \frac{1}{6}x^2+\frac{-5}{18}x+\frac{19}{108}$, and the general solution is $y(x) = y_c(x) + y_p(x) = c_1e^{-2x} + c_2e^{-3x} + \frac{1}{6}x^2+\frac{-5}{18}x+\frac{19}{108}$.