Finding Particular Solutions Inhomogeneous ODE's

By failing to prepare, you are preparing to fail, Benjamin Franklin.

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.

Finding Particular Solutions for Inhomogeneous ODE’s

In this discussion, we will explore methods for finding particular solutions to second-order linear inhomogeneous differential equations with constant coefficients. Specifically, we will focus on equations of the form: y’’ + Ay’ + By = f(x)

This equation is inhomogeneous because of the non-zero forcing or inhomogeneous term f(x). Our goal is to find a particular solution y_p to this equation. Once whe have y^p(t), the general solution to the ODE is the sum of the particular solution and the complementary solution y_c(x), which solves the associated homogeneous equation y’’ + Ay’ + By = f(x).

Hence, the general solution y(x) = y_p(x) + y_c(x) = y_p + c₁y₁ + c₂y₂ where c₁y₁ + c₂y₂ is the complementary or homogeneous solution.

We are particularly interested in finding y_p when the forcing term f(x) takes certain common form, such as:

• Exponential Functions: f(x) = e^ax, where a is a constant.
• Trigonometric functions: f(x) = sin(wx) or f(x) = cos(wx), where w is a constant.
• Complex exponentials: f(x) = e^(a+iw)x = e^αx, where α = a + iw is a complex number.

To make the process clearer, we rewrite the original ODE using differential operator notation. Let D represent the derivative with respect to x. The original ODE: y’’ + Ay’ + By = f(x) becomes (D² + AD + B)y = f(x).

We can then introduce the operator p(D) defined as: p(D) = D² + AD + B.

p(D) can be understood as a linear operator (it acts on functions by applying derivatives and combining terms linearly) and also a formal polynomial in D (it behaves like a polynomial where D is treated as an algebraic symbol).

This allows us to rewrite the ODE as: p(D)y = f(x)

The substitution Rule. A key result for solving such equations is the Substitution Rule for exponentials. This rule states that: p(D)e^αx = p(α)e^αx where α is any complex number.

Proof:

Apply the differential operator to the exponential function e^αx:

p(D)e^αx = (D² + AD + B)e^αx =[By linearity, calculate each of them] D²e^αx + ADe^αx + Be^αx = α²e^αx + Aαe^αx + Be^αx =[Factor out e^ax] (α² +Aα + B)e^αx = p(α)e^αx ⇒ p(D)e^αx = p(α)e^αx∎

Exponential input theorem. It states that if f(x) = e^ax, the inhomogeneous ODE: y'' + Ay' + By = e^αx has a particular solution of the form $y_p = \frac{e^{αx}}{p(α)}$ assuming that p(α) ≠ 0.

Proof.

Start with the equation p(D)y_p = e^αx and assume $y_p = \frac{e^{αx}}{p(α)}$

Substituting this into the equation:

$p(D)y_p = p(D)\frac{e^{αx}}{p(α)} =[\text{Use the substitution Rule}] \frac{p(α)e^{αx}}{p(α)} = e^{αx}$ ∎

Example: Finding a Particular Solution with a Complex Forcing Term.

Find the particular solution for the equation y’’ -y’ +2y = 10e^-xsin(x)

Step 1: Convert to Complex Form To handle the trigonometric function sin(x), express it as the imaginary part of the complex exponential e^ix. Therefore, the forcing term becomes: f(x) = 10e^-xsin(x) = $Im(10e^{-x}e^{ix}) = Im(10e^{(-1+i)x})$

Let $\tilde{y}(x)$ be the complex solution to the ODE: $(D^2-D+2)\tilde{y} = 10e^{(-1+i)x}$

Step 2: Apply the Exponential Input Theorem

The characteristic polynomial is p(α) = α² - α + 2. Substituting α = −1+i: $p(-1+i) = (-1+i)^2-(-1+i)+2 = 3 -3i$

Thus, the complex particular solution is: $\tilde{y_p} = \frac{e^{αx}}{p(α)} = \frac{10e^{(-1+i)x}}{3 -3i} = \frac{10e^{(-1+i)x}}{3(1 -i)} = \frac{10}{3}\frac{1+i}{(1-i)(1+i)}e^{(-1+i)x} = \frac{10}{3}\frac{1+i}{2}e^{-x}(cos(x)+isin(x)) = \frac{5}{3}(1+i)e^{-x}(cos(x)+isin(x))$

Extract the imaginary part:

$y_p = Im(\tilde{y_p}) = \frac{5}{3}e^{-x}(cos(x)+sin(x)) =[\text{Alternatively, we can express this as:}] \frac{5}{3}e^{-x}\sqrt{2}cos(x-\frac{π}{4})$

Recall: Acos(x) + Bsin(x) = Rcos(x -ϕ) where A = B = 1, R is the amplitude, R = $\sqrt{1+1}=\sqrt{2}$ and ϕ is the phase shift, cos(ϕ) = sin(ϕ) = ¹⁄_√2. Since both the sine and cosine of ϕ are equal, we recognize that: ϕ = arctan(¹⁄₁) = ^π⁄₄

Special case: p(α) = 0

When p(α)=0, this simple approach using the Exponential Input Theorem fails. This occurs when α is a root of the characteristic equation. In this case, we must modify the form of the particular solution.

Exponential-shift rule. If p(α) = 0, it states that: p(D)e^axu(x) = e^axp(D +a)u(x).

Proof:

Particular case p(D) = D

p(D)e^axu(x) =[Particular case p(D) = D] De^axu(x) = [Product rule] e^axu’ + ae^axu =[Factor e^ax] e^ax(u’ + au) = e^ax(D + a)u = e^axp(D + a)u(x)

p(D) = D²

D²e^axu = D(De^axu) =[Previous result] D(e^ax(D + a)u) = [Previous result, but u is in this case (D + a)u] e^ax(D + a)(D + a)u = e^ax(D + a)²u.

This result is derived from the product rule of differentiation and can be applied iteratively to higher-order terms.

Let y’’ + Ay’ + By = e^ax be an inhomogeneous ODE where a is a constant. Then, a particular solution is given by $y_p = \frac{xe^{ax}}{p’(a)}$ assuming that p(a) = 0 and a is a simple root of the characteristic equation p(D) = D² + AD + B. If a is a double root, the particular solution is $y_p = \frac{x^2e^{ax}}{p’’(a)}$

Proof (Simple root case)

If a is a simple root of p(D), then p(D) = D² +AD + B. If a is a root of this characteristic equation, we know that p(a) = a² + Aa + B = 0. In the case of a simple root, we differentiate p(D) with respect to D, p’(D) = 2D + A. Substituting D = a, we get, p’(a) = 2a + A.

Establish the claim: p(D)y_p = $p(D)\frac{e^{ax}·x}{p’(a)} = e^{ax}$

$p(D)y_p = p(D)\frac{e^{ax}·x}{p’(a)}$

Now, use the Exponential Shift Rule, which states that: p(D)e^axu(x) = e^axp(D+a)u(x), and apply this rule to e^ax·x where u(x) = x, so

$p(D)y_p = p(D)\frac{e^{ax}·x}{p’(a)} = \frac{e^{ax}}{p’(α)}·p(D+a)x$

Next, calculate p(D + a) ↭[p(D) = D² + AD + B] (D + a)² + A(D + a) + B² = D² + 2aD + a² + AD + Aa + B =[a is a root, a² +Aa + B = 0] D² +(2a + A)D

Now, applying this to x, we calculate: p(D + a)x = (D² +(2a + A)D)x = D²(x) + (2a + A)D(x) =[D(x) = 1, D’’(x) = 0] 0 + (2a + A)·1 = (2a + A) = p’(a)

The operator D is shorthand for “taking the derivative with respect to x”, D(x) is 1, D(1) = 0 because the derivative of a constant is zero, D²(x) = 0.

$p(D)y_p = p(D)\frac{e^{ax}·x}{p’(a)} = \frac{e^{ax}}{p’(a)}p(D+a)x = \frac{e^{ax}}{p’(a)}p’(a) = e^{ax}$∎

• Find a particular solution to the inhomogeneous second-order linear differential equation y’’ -3y’ + 2y = e^x.

Our goal is to find a specific function y_p(x) that satisfies this equation. Once we have the particular solution, the general solution to the ODE can be written as the sum of the complementary (homogeneous) solution and the particular solution: y(x) = y_c(x) + y_p(x).

To apply the theorem, we first need to determine the characteristic equation associated with the homogeneous part of the ODE: y’’ -3y’ + 2y = e^x. We assume solutions of the form y = e^rx, where r is a constant to be determined.

Substituting y = e^rx into the homogeneous equation yields: $r^2e^{rx}-3re^{rx}+2e^{rx} = 0$⇒[Divide both sides by e^rx (since e^rx > 0)] gives the characteristic equation associated with the homogeneous ODE: r² -3r + 2 = 0.

Their roots are r₁ = 1, r₂ = 2. Since we have two distinct real roots, the general solution to the homogeneous equation is: $y_c(x)=c_1e^{r_1x}+c_2e^{r_2x}=c_1e^{x}+c_2e^{2x}$ where c₁ and c₂ are arbitrary constants determined by initial conditions.

The characteristic polynomial is obtained by replacing y′′ with D², y′ with D, and y with 1, respectively where D is the differential operator: p(D) = D² -3D +2 = 0.

Solve the characteristic equation D² -3D +2 =[Factoring the quadratic equation gives:] (D -1)(D -2), so the roots are D = 1 and D = 2. These roots corresponds to the solutions e^x and e^2x for the homogeneous ODE y’’ -3y’ + 2y = 0.

Since the right-hand side of the inhomogeneous ODE is e^x, we look at the root D = 1. 1 is a simple root of the characteristic equation because it only appears once in the factorization (D−1)(D−2).

Using the Exponential Shift Method

• If f(x) = e^ax and a is a simple root of the characteristic equation, then we try a particular solution of the form: y_p(x) = xe^ax·constant.
• If a is a repeated root (multiplicity m), then we try: y_p(x) = x^me^ax·polynomial in x.

By the theorem, when the right-hand side of the ODE is of the form e^ax and a is a simple root of the characteristic equation, the particular solution is given by: $y_p = \frac{xe^{ax}}{p'(a)} = \frac{xe^x}{p'(1)}$

The characteristic polynomial is: p(D) = D² -3D +2. Differentiate this with respect to D: p’(D) = 2D -3. Now, substitute D = 1: p’(1) = 2 -3 = -1.

y_p(x) $= \frac{xe^x}{p’(1)} = \frac{xe^x}{-1} = -xe^x$, and the general solution = y(x) = y_c(x) + y_p(x) = $c_1e^{x}+c_2e^{2x}-xe^x$ where c₁ and c₂ are arbitrary constants determined by initial conditions.

Resonance in Differential Equations

We are exploring the phenomenon of resonance through the process of finding particular solutions to second-order linear inhomogeneous differential equations of the form: y’’ + w₀²y = cos(w₁t).

This equation describes a system (such as a mass-spring system or a pendulum) that experiences external forcing, represented by cos(w₁t) where:

• y = y(t) is the unknown function of time t.
• y’ and y’’ are the first and second derivatives of y with respect to t, respectively.
• w₀² is the natural frequency of the system (how it naturally oscillates without external forces).
• w₁ is forcing frequency, representing an external periodic input, like pushing the pendulum at regular intervals.

We assume w₁ ≠ w₀, meaning the input frequency is not exactly the same as the system’s natural frequency. The general solution to the inhomogeneous differential equation consists of two parts:

1. Complementary Solution (y_c): Solves the associated homogeneous equation: y’’+ w₀²y = 0.
2. Particular Solution (y_p). A specific solution that accounts for the inhomogeneous part cos(w₁t).

Thus, the general solution is: y(t) = y_c(t) + y_p(t).

Assuming a solution of the form y = e^rt, we substitute into the homogeneous equation: $r^2e^{rt}+w_0^2e^{rt} = 0$ ⇒[Divide both sides by e^rt (since e^rt ≠ 0)] r² + w₀² = 0 ⇒[Solve for r] r = ±iw₀ (pure imaginary). Therefore, the complementary solution is: y_c = c₁cos(w₀t) + c₂sin(w₀t) where c₁ and c₂ are arbitrary constants determined by initial conditions. This describes the natural oscillatory behavior of the system with frequency w₀.

Finding the particular solution y_p(t).

Convert the Equation to Differential Operator Form: First, we rewrite the differential equation using D, where D = d/dt represents the derivative with respect to t. So: (D² + w₀²)y = cos(w₁t).

This is a second-order linear inhomogeneous differential equation.

Switch to Complex Exponentials (Easier to Solve): Since working with cosine functions can be tricky, we switch to using complex exponentials, as they simplify the process. Using Euler’s formula, we know that: $cos(w_1t)=Re(e^{iw_1t})$

Thus, solving the equation with $e^{iw_1t}$ will give us a complex solution, and we’ll take the real part at the end to get our solution for the cosine function.

The equation becomes: $(D^2+w_0^2)\tilde{y} = e^{iw_1t}$ where $\tilde{y}$ is the complex solution, and we are only interested in the real part of $\tilde{y}$ at the end.

To solve for $\tilde{y_p}$, we use the fact that e^iw₁t is a simple exponential, and we can use the [Exponential input theorem. It states that if f(x) = e^ax, the inhomogeneous ODE: y'' + Ay' + By = e^αx has a particular solution of the form $y_p = \frac{e^{αx}}{p(α)}$ assuming that p(α) ≠ 0.]

p(D) is the characteristic polynomial of the differential operator D² + w₀². p(iw₁) = $(iw_1)^2+w_0^2 = w_0^2-w_1^2$. Since $w_0^2-w_1^2 ≠ 0$ (non-resonant case), we proceed.

Thus, the particular solution becomes:

$\tilde{y_p} = \frac{e^{iw_1t}}{(iw_1t)^2+w_0^2} =[\text{Simplifying the denominator:}] \frac{e^{iw_1t}}{w_0^2-w_1^2}$

Now, we take the real part of $y_p$, to find the solution for y_p: $y_p = Re(\tilde{y_p}) = \frac{cos(w_1t)}{w_0^2-w_1^2}$. This is the particular solution when the input frequency is not equal to the system’s natural frequency, w₁ ≠ w₀

Resonance (When w₁ ≈ w₀):

Resonance occurs when the input frequency w₁ is close to the natural frequency w₀ of the system. In such cases, the response of the system becomes amplified (the denominator $w_0^2-w_1^2$ get closer to zero, which leads to the amplitude of the response growing significantly), as the forcing frequency aligns with the system’s natural oscillation (Refer to Figure A for a visual representation and aid in understanding it)

Special Case: Resonance (When w₀ ≈ w₁)

If w₀ ≈ w₁, we have a resonance condition. In this case, the earlier approach doesn’t work because the denominator becomes zero, meaning the solution must be treated differently.

We start with (D² + w₀²)y = cos(w₁t).

Again, we convert to complex exponentials:

$(D^2+w_0^2)\tilde{y} = e^{iw_0t}$ but iw₀ is a root of the characteristic polynomial, $(D^2+w_0^2)(iw_0) = (iw_0)^2+w_0^2 = 0$

To find a particular solution [$y_p = \frac{xe^{ax}}{p’(a)}, p’(D) = 2D$], $p’(iw_0) = 2iw_0$ $\tilde{y_p} = \frac{te^{iw_0t}}{2iw_0}=\frac{tcos(w_0t)+itsin(w_0t)}{2iw_0} = \frac{cos(w_0t)ti-sin(w_0t)t}{-2w_0}$

Finally, we take the real part of this to get the solution for the cosine case:

$y_p = Re(\tilde{y_p}) = \frac{tsin(w_0t)}{2w_0}$. When w₁ ≈ w₀, the solution has a steadily increasing amplitude. In particular:

• The term tsin(w₀t) shows that the amplitude grows linearly with time.
• This is a characteristic of resonance: the system’s response builds over time, leading to larger and larger oscillations (Refer to Figure C for a visual representation and aid in understanding it).

  If there’s no damping or limit to the oscillation, resonance can cause the amplitude to grow indefinitely, which is why resonance can be both powerful and dangerous (e.g., bridges collapsing due to resonance from wind or foot traffic)

We have previously found a particular solution: $y_p = \frac{cos(w_1t)}{w_0^2-w_1^2}$. This is the particular solution when the input frequency is not equal to the system’s natural frequency, w₀ ≠ w₁

Special Case: Resonance (w₁ ≈ w₀)

When the forcing frequency w₁ approaches the natural frequency w₀, the system experiences resonance. In this case, the particular solution $y_p = \frac{cos(w_1t)}{w_0^2-w_1^2}$ becomes problematic as the denominator approaches zero, leading to a singularity.

To deal with this, we find another particular solution that takes into account the resonance condition. We start with the complementary solution: $y_c = -\frac{cos(w_0t)}{w_0^2-w_1^2}$ where $c_1=\frac{-1}{w_0^2-w_1^2},c_2=0$

So another particular solution is $\frac{cos(w_1t)}{w_0^2-w_1^2} -\frac{cos(w_0t)}{w_0^2-w_1^2}$

Taking the limit as w₁ → w₀:

$\lim_{w_1 \to w_0} \frac{cos(w_1t)}{w_0^2-w_1^2} -\frac{cos(w_0t)}{w_0^2-w_1^2} = \lim_{w_1 \to w_0} \frac{cos(w_1t)-cos(w_0t)}{w_0^2-w_1^2} $[L’Hopital Rule, please note the variable is w₁] $\lim_{w_1 \to w_0} \frac{-sin(w_1t)·t}{-2·w_1} = \frac{sin(w_0t)·t}{2w_0}$ This result is the particular solution when w₁ = w₀, which corresponds to the resonance case.

This solution grows linearly in time (t), reflecting the amplification of oscillations characteristic of resonance. The amplitude of the oscillation increases over time without bound, which can lead to destructive consequences if no damping is present.

Geometric Interpretation of the Resonance Condition (w₁ ≈ w₀)

We can also provide a geometric interpretation of the behavior as w₁ approaches w₀.

Consider the difference:

$\frac{cos(w_1t)-cos(w_0t)}{w_0^2-w_1^2}= [\text{Using the identity trigonometry: }cosB -cosA = 2sin(\frac{A-B}{2})sin(\frac{A+B}{2})] = \frac{2sin(\frac{w_0-w_1}{2})t+sin(\frac{w_0+w_1}{2})t}{w_0^2-w_1^2}$ This equation consists of two parts: (Refer to Figure B for a visual representation and aid in understanding it)

• $\frac{2sin(\frac{w_0-w_1}{2})t}{w_0^2-w_1^2}$. This represents oscillations with very small frequency (since w₁-w₀ is small). The period of this oscillation is large. Because of the term $\frac{2}{w_0^2-w_1^2}$, the motion increases at a steadily increasing amplitude as w₁ approaches w₀.
• $\frac{sin(\frac{w_0+w_1}{2})t}{w_0^2-w_1^2} ≈ \frac{sin(w_0)t}{w_0^2-w_1^2}$, this is a pure oscillatory term with frequency close to w₀.

As w₁ approaches w₀, the system exhibits a phenomenon known as beats, where the amplitude of the oscillations grows slowly over time. At exact resonance (w₁ = w₀), the oscillation amplitude grows linearly over time, leading to increasingly large oscillations. This growth can continue indefinitely in undamped systems, resulting in increasingly large oscillations that may cause structural damage, failure or other dangerous effects.

Why Resonance Matters:

• Engineering Applications: Structures and mechanical systems must be designed to avoid resonance at operating frequencies.
• Safety Concerns: Unchecked resonance can lead to catastrophic failures (e.g., Tacoma Narrows Bridge collapse).
• Control Systems: Understanding resonance helps in designing damping mechanisms to mitigate its effects.

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