Integrating Factors for Linear Differential Equations

Now I Am Become Death, the Destroyer of Worlds, Oppenheimer.

Do, or do not. There is no try, Yoda.

Recall. Algebraic and differential equations.

An algebraic equation is a mathematical statement asserting that two algebraic expressions are equal.. These expressions are constructed using:

1. Dependent and independent variables. Variables are symbols representing unknown quantities. Variables can be:
   Independent Variables: Variables or input that can be chosen (or change) freely.
   Dependent Variables: Variables that depend on the independent variables.
2. Constants. Fixed numerical values that do not change, e.g., 2, π, 7.
3. Algebraic operations. Operations such as addition, subtraction, multiplication, division, exponentiation, powers, and root.

An algebraic equation typically has the following general form: Expression₁ = Expression₂, where the two expressions are set equal to each other.

Examples: y = 2x + 4 (simple linear equation, represents a straight line with a slope of 2 and a y-intercept of 4), ax² + bx + c = 0 (quadratic equation, it represents a parabola), $ln(xy + t) = z·sin(\sqrt{x})$.

Definition. A differential equation is an equation that involves one or more functions and their derivatives. It relates the function itself (the dependent variable), its derivatives with respect to one or more independent variables (rates of change), and the independent variable(s) themselves, e.g., $\frac{dy}{dx} = 3x +5y, y' + y = 4xcos(2x), \frac{dy}{dx} = x^2y+y, etc.$

If an algebraic equation is a “snapshot” of a relationship, a differential equation is the “instruction manual” for how that relationship moves and changes.

Key Components (e.g., $\frac{dy}{dx} = 3x +5y$):

• Dependent variables: The function y (or y(x)) we are trying to find or solve for. Variable(s) (y) that depend on one or more other variables.
• Independent variables: Variable(s) (x) upon which the dependent variables depend and we differentiate with respect to (often time or space).
• Derivatives: Expressions like $\frac{dy}{dx}$ or y’ that represent the rate at which the dependent variable (y) change with respect to the independent variable (x).

Separation of Variables

Separable differential equations are a special class of differential equations that can be manipulated algebraically to separate the variables, allowing us to integrate and find solutions.

A separable differential equation is any differential equation that can be written in the form $\frac{dy}{dx}=f(x)g(y)$. This form allows us to separate the variables x and y on opposite sides of the equation, facilitating integration.

Steps to Solve Separable Differential Equations

1. Check for any values of y that make g(y) = 0. These correspond to constant (equilibrium) solutions where $\frac{dy}{dx} = 0$.
2. Rewrite the equation to isolate y terms on one side and x terms on the other: $\frac{dy}{g(y)} = f(x)dx$
3. Integrate both sides of the equation: $\int \frac{dy}{g(y)} = \int f(x)dx$
4. Solve the resulting equation for y if possible to find the general solution.
5. If an initial condition (e.g., x₀ = y₀) is provided, substitute x₀ and y₀ into the general solution and solve for the constant of integration C.

Differential Equations: Integral Factors

Definition. A first-order differential equation is called linear if it can be written in the standard form $\frac{dy}{dx} +P(x)y = Q(x)$ where P(x) and Q(x) are continuous functions of x over some interval I.

This equation is considered “first-order” because it only involves the first derivative of y with respect to x. It is considered “linear” because the dependent variable y and its derivative y′ both appear to the first power and are not multiplied together or nested inside other functions.

Examples of Linear First-Order Differential Equations

These equations can all be manipulated to fit the standard linear form $\frac{dy}{dx} +P(x)y = Q(x)$:

• $y' + 2xy = \sin(x)$ (it is already in standard form with P(x) = 2x and Q(x) = sin(x)).
• $e^x y' - \tan(x)y = 3x$ (it can be divided by $e^x$ to reach standard form).
• $(x^2 + 7)y' - 3\cos(x) = xy$ (it can be rearranged to $(x^2 + 7)y' - xy = 3\cos(x)$ and then divided by $(x^2 + 7)$).

Non-Examples (Nonlinear Equations)

These equations are not linear due to the presence of non-linear terms involving $y$ or $y'$:

• $y' + y^2 = \tan(x)$ (The y² term makes it nonlinear.)
• $y' \cdot y = e^x$ (The multiplication of the derivative $y'$ by the variable $y$ makes it nonlinear.)
• $e^y + 2y' = 0$ (The $e^y$ term introduces a nonlinear exponential relationship).

The Method of Integrating Factors

The integrating factor method is a powerful analytical technique for solving first-order linear ordinary differential equations (ODEs).

This method involves multiplying the entire equation by a carefully chosen function, called the integrating factor. The purpose of this function, denoted μ(x), is to transform the left side of the equation into the exact derivative of a product, which simplifies integration.

Deriving the Integrating Factor

Let’s start with our equation in standard form and multiply it by an unknown positive function $\mu(x)$: $ \mu(x)\frac{dy}{dx} + \mu(x)P(x)y = \mu(x)Q(x)$

We want to force the left side of this equation to be the derivative of the product $(\mu(x)y)$. According to the product rule, the derivative of $\mu(x)y$ is: $ \frac{d}{dx}(\mu(x)y) = \mu(x)\frac{dy}{dx} + \mu'(x)y $

For our forced equation to match this product rule expansion, the $y$ terms must be equal. Therefore, we require: $ \mu'(x) = \mu(x)P(x)$

This gives us a separable differential equation for $\mu(x)$:

1. Separate variables: $\frac{\mu'(x)}{\mu(x)} = P(x)$
2. Integrate both sides: $\int \frac{1}{\mu} d\mu = \int P(x)dx \implies \ln|\mu| = \int P(x)dx$
3. Take the exponential of both sides: $\mu(x) = e^{\int P(x)dx}$

(Note: We do not need to include an arbitrary constant of integration $C$ here because any single valid integrating factor will work to solve the original ODE).

By multiplying our original ODE by this specific $\mu(x)$, the equation collapses into: $ \frac{d}{dx}(\mu(x) \cdot y) = \mu(x) \cdot Q(x) $ (💡 Recognize the left side as the derivative of a product.)

From here, we simply integrate both sides with respect to $x$ to find the solution.

The Method Step-by-Step

1. Ensure that the differential equation is written in the standard linear form: $\frac{dy}{dx} +P(x)y = Q(x)$ (If $y'$ has a coefficient, divide the entire equation by that coefficient first).
2. Calculate the Integrating Factor. The integral factor μ(x) is given by: $μ(x) = e^{\int P(x)dx}$
3. Multiply both sides of the differential equation by the integrating Factor μ(x): $μ(x)\frac{dy}{dx} +μ(x)P(x)y = μ(x)Q(x).$ By construction, the left-hand side becomes the derivate of the product μ(x)y with respect to x, i.e., $\frac{d}{dx}(μ(x)y) = μ(x)Q(x).$
4. Integrate both sides of the equation: $\int \frac{d}{dx}(μ(x)·y)dx = \int μ(x)Q(x)dx ↭[\text{Simplifying, this becomes:}] μ(x)y = \int μ(x)Q(x)dx + C$ where C is the constant of integration.
5. Solve for y. Finally, divide both sides by μ(x) to isolate the general solution y(x): y = $\frac{1}{μ(x)}(\int μ(x)Q(x)dx + C)$

Examples

• $\frac{dy}{dx} -3y = 6$.

1. The equation is already in standard form: $\frac{dy}{dx} +P(x)y = Q(x)$ where P(x) = -3 and Q(x) = 6.
2. Second, we compute the integrating factor, μ(x). It is defined as $μ(x) = e^{\int P(x)dx} = e^{\int -3dx} = e^{-3x}$
3. Third, we write the differential equation in the form $\frac{d}{dx}(μ·y)=μ·Q(x)$ by multiplying both sides of the differential equation by the integrating factor, $e^{-3x}(\frac{dy}{dx} -3y) = 6·e^{-3x} ↭ \frac{d}{dx}(e^{-3x}·y) = 6·e^{-3x}$.
4. Next, we can solve the differential equation integrating with respect to x, $e^{-3x}·y = \int 6·e^{-3x}dx$ =[u = -3x, du = -3dx ⇒ $dx = \frac{-du}{3}$] $\int -2e^udu = -2e^u = -2e^{-3x} + C ⇒ e^{-3x}·y = -2e^{-3x} + C$ where C is the constant of integration.
5. Finally, solving for y by multiplying by e^3x, we get y = $-2 + C·e^{3x}$

• y’ + y = 5x².

1. Verify Standard Form. The equation is already in standard form: $\frac{dy}{dx} +P(x)y = Q(x)$ where P(x) = 1 and Q(x) = $Q(x) = 5x^2$.

2. Second, we compute the integrating factor, $\mu(x)$. It is defined as $μ(x) = e^{\int P(x)dx} = e^{\int dx} = e^{x}$.
   Recall that $\frac{d}{dx}(\mu \cdot y) = \mu y' + \mu' y$.

3. Third, multiply both sides of the equation by the integrating factor: $e^xy' + e^xy = e^x5x^2$ ↭[Recall $\frac{d}{dx}(μ·y)=μ·Q(x)$, the left-hand side is the derivative of the product μ(x)=e^x and y] $\frac{d}{dx}(e^x·y) = e^x5x^2$

4. Next, we can solve the differential equation integrating with respect to x, $\int \frac{d}{dx}(e^x·y)dx = \int e^x5x^2dx ↭ e^x·y = \int e^x5x^2dx$

   $\int e^x5x^2dx = $[Integration by parts ∫udv=uv−∫vdu, u = 5x² ⇒ du = 10xdx; dv = e^xdx ⇒ v = e^x] = $5x^2e^x -\int 10xe^xdx$ [Integration by parts ∫udv=uv−∫vdu, u = 10x ⇒ du = 10dx; dv = e^xdx ⇒ v = e^x] $5x^2e^x -(10xe^x-\int 10e^xdx) = 5x^2e^x-10xe^x +10e^x + C$

5. Therefore, $e^x·y = \int e^x5x^2dx = 5x^2e^x-10xe^x +10e^x + C ⇒[\text{Divide both sides by eˣ to solve for y}] y = 5x^2-10x+10 + Ce^{-x}$

• $\frac{dy}{dx} + 2y = x^2.$

1. Verify Standard Form. The equation is already in standard form: $\frac{dy}{dx} +P(x)y = Q(x)$ where P(x) = 2 and Q(x) = x².

2. Second, we compute the Integrating Factor $\mu(x)$. It is defined as $μ(x) = e^{\int P(x)dx} = e^{\int 2dx} = e^{2x}$

3. Third, we write the differential equation in the form $\frac{d}{dx}(μ·y)=μ·Q(x)$ by multiplying both sides of the differential equation by the integrating factor, $e^{2x}(\frac{dy}{dx} + 2y) = e^{2x}·x^2 ↭ \frac{d}{dx}(e^{2x}·y) = e^{2x}·x^2$.

4. Next, we can solve the differential equation by integrating with respect to x, $e^{2x}·y = \int e^{2x}·x^2 dx$ =[Recall Integration by parts formula $\int udv = uv - \int vdu$, u = x² (so du = 2xdx) dv = $e^{2x}dx$, so v = $\frac{1}{2}e^{2x}$] $x^2\frac{1}{2}e^{2x}- \int \frac{1}{2}e^{2x}2xdx = \frac{1}{2}x^2e^{2x}- \int e^{2x}xdx$

   $\int e^{2x}xdx$ =[Apply integration by parts again, u = x, dv = $e^{2x}dx, du = dx, v = \frac{1}{2}e^{2x}$] $\frac{1}{2}xe^{2x} -\int \frac{1}{2}e^{2x}dx = \frac{1}{2}xe^{2x} -\frac{1}{4}e^{2x} + C.$
   Substitute the result back into our equation: $e^{2x} \cdot y = e^{2x}\left(\frac{1}{2}x^2 - \frac{1}{2}x + \frac{1}{4}\right) - C$

5. Finally, we can solve for y: $y = \frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{4} - C·e^{-2x}$ where C is a constant determined by initial conditions, if provided.

• Newton’s Law of Cooling $\frac{dT}{dt}+kT = kT_e$.

This differential equation models processes like Newton’s Law of Cooling, where T is the temperature of an object over time, k is a positive constant (cooling rate), and T_e is the surrounding environment’s temperature.

1. Write the differential equation in standard form: $\frac{dT}{dt}+kT = kT_e$. It is already in standard form: $\frac{dT}{dt} +P(t)T = Q(t)$ where P(t) = k and Q(t) = kT_e.
2. Compute the integrating factor: $μ = e^{\int P(t)dt} = e^{\int kdt} = e^{kt}$.
3. Multiply both sides of the differential equation by the integrating factor e^kt: $e^{kt}\frac{dT}{dt}+kTe^{kt} = kT_ee^{kt}$
4. By construction, the left-hand side becomes the derivative of the product e^ktT, i.e., $(e^{kt}T)' = kT_ee^{kt}$
5. Integrate both sides of the equation with respect to t: $e^{kt}T = \int kT_ee^{kt}dt + C$ where C is the constant of integration.
6. Solve for T(t), $T(t) = e^{-kt}\int kT_e(t)e^{kt}dt + Ce^{-kt}$

Scenario A: The Environment Temperature varies over time, $T_e(t)$

If the external temperature changes over time, we integrate from time $t=0$ to some arbitrary time $t$. To avoid confusing variables, we use a “dummy variable” $\tau$ for the integration: $\int_0^t \frac{d}{d\tau}(e^{k\tau}T(\tau)) \, d\tau = \int_0^t k T_e(\tau)e^{k\tau} \, d\tau$

By the Fundamental Theorem of Calculus: $e^{kt}T(t) - e^0T(0) = \int_0^t k T_e(\tau)e^{k\tau} \, d\tau$

Let the initial temperature $T(0) = T_0$. Then, $e^{kt}T(t) - T_0 = \int_0^t k T_e(\tau)e^{k\tau} \, d\tau$

Solve for $T(t)$: $\boxed{T(t) = e^{-kt}\int_0^t k T_e(\tau)e^{k\tau} \, d\tau + T_0e^{-kt}}$

Physical interpretation:
The term $T_0e^{-kt}$ is the transient response (how the object “forgets” its initial temperature over time).
Forced/steady-state term: $e^{-kt}\int_0^t k T_e(\tau)e^{k\tau} \, d\tau$.

Scenario B: The Environment Temperature is a Constant, $T_e$

If the external temperature is constant, solving the indefinite integral is much simpler.

$\frac{d}{dt}(e^{kt}T) = kT_e e^{kt}$. Integrate both sides: $e^{kt}T = \int kT_e e^{kt} \, dt = kT_e \left(\frac{1}{k}e^{kt}\right) + C$

$e^{kt}T = T_e e^{kt} + C$. Divide by $e^{kt}$: $T(t) = T_e + Ce^{-kt}$

To find $C$, apply the initial condition $T(0) = T_0$: $T(0) = T_e + Ce^0 \implies T_0 = T_e + C \implies C = T_0 - T_e$

Final Solution: The solution to the differential equation modeling Newton’s Law of Cooling for a constant external temperature $T_e$ and initial condition $T(0) = T_0$ is: $\boxed{T(t) = T_e + (T_0 - T_e)e^{-kt}}$

Physical Interpretation: This elegantly describes the temperature $T(t)$ decaying exponentially from its initial state $T_0$ toward the room temperature $T_e$ as $t \to \infty$. The speed of this decay is entirely determined by the constant $k$.

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