Integral Factors

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

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

Recall

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

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

• Dependent variables: Variable(s) (y) that depend on one or more other variables. The function y (or y(x)) we are trying to solve for.
• Independent variables: Variable(s) (x) upon which the dependent variables depend.
• 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 y(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 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 form only involves the first derivative of y with respect to x, hence it is called a first-order differential equation.

Examples of Linear First-Order Differential Equations

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

• y’ + 2xy = sin(x)
• e^xy’ -tan(x)y = 3x
• (x² + 7)y’ -3cos(x) = xy

Non-Examples (Nonlinear Equations)

These equations are not linear due to the presence of non-linear terms in y or y′:

• y’ + y² = tan(x) (The y² term makes it nonlinear.)
• y’·y = e^x (The multiplication of y’ and y makes it nonlinear.)
• e^y + 2y’ = 0 (The e^y term introduces a nonlinear relationship between y and y’).

The Method of Integrating Factors

The integrating factor method is a useful 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 derivative of a product, which simplifies integration.

First, we multiply our equation for the integrating factor: uy’ + P(x)u(x)y = uQ(x)

We aim to get (uy)’ = uy’ + P(x)uy and this only works if u’y = P(x)uy ↭ u’ = P(x)u ⇒[Separate variables] $\frac{u’}{u} = P ⇒[\text{Taking integrals}] \int \frac{du}{u} = \int P(x)dx ↭ lnu = \int P(x)dx ⇒[\text{Taking exponentials}] u = e^{\int P(x)dx}$

We are not going to use any arbitrary constant since only one integrating factor u is required. The integrating factor, μ(x), is defined as $μ(x) = e^{\int P(x)dx}$

The solution to the differential equation is obtained by multiplying both sides of the equation by the integrating factor $μ(\frac{dy}{dx} +P(x)y) = μ·Q(x)$, which becomes $\frac{d}{dx}(μ·y)=μ·Q(x)$ (💡Recognize the left side as the derivative of a product.) and then integrating both sides with respect to x (This transformation allows us to integrate both sides easily and find the solution), $μ·y = \int μ·Q(x)dx$, and now the ODE can be solved.

The Method Step-by-Step

1. Ensure that the differential equation is written in the standard linear form: $\frac{dy}{dx} +P(x)y = Q(x).$
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 get 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. The equation is already in standard form: $\frac{dy}{dx} +P(x)y = Q(x)$ where P(x) = 1 and Q(x) = 5x².

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

3. Third, multiply both sides of the equation by the integrating factors: $e^xy’ + e^xy = e^x5x^2$ ↭[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. 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, μ(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 = $e^{2x}dx, du = dx, v = \frac{1}{2}e^{2x}$] $\frac{1}{2}xe^{2x} -\int \frac{1}{2}e^{2x}dx = \frac{1}{2}xe^{2x} -\frac{1}{4}e^{2x} + C.$ Therefore, $e^{2x}·y = \frac{1}{2}x^2e^{2x}- \int e^{2x}xdx = \frac{1}{2}x^2e^{2x} -\frac{1}{2}xe^{2x} +\frac{1}{4}e^{2x} - C = e^{2x}(\frac{1}{2}x^2-\frac{1}{2}x+\frac{1}{4}) - C$

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.

• Solve the First-Order Linear Differential $\frac{dT}{dt}+kT = kT_e$.

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, 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, $T(t) = e^{-kt}\int kT_e(t)e^{kt}dt + Ce^{-kt}$

If an initial condition T(0) = T₀ is provided, we can determine C: $T_0 = e^{-0·t}\int_{0}^{0} kT_e(t)e^{kt}dt + C = C$. Thus, C = T₀ and $T(t) = e^{-kt}\int_{0}^{t} kT_e(t_1)e^{kt_1}dt_1 + T_0e^{-kt}$.

If T_e(t) is bounded and k > 0, the exponential term $T_0e^{-kt}$ approaches zero as t → ∞, so $e^{-kt}\int_{0}^{t} kT_e(t_1)e^{kt_1}dt_1$ is the steady-state solution.

For constant external temperature T_e(t), T_e(t) = T_e, $T = e^{-kt}\int kT_e(t)e^{kt}dt + Ce^{-kt} = e^{-kt}kT_e\int e^{kt}dt + Ce^{-kt} = e^{-kt}kT_e·\frac{e^{kt}}{k} + Ce^{-kt} = T_e + Ce^{-kt}=$ [Using the initial condition T(0) = T₀] T(0) = T₀ = T_e + C⇒ C = T₀ - T_e⇒ The solution to the differential equation modelling Newton's Law of Cooling for constant external temperature T_e and initial condition T(0) = T₀ is $T = T_e + (T_0-T_e)e^{-kt}$.

This solution describes the temperature T(t) approaching T_e as t → ∞, with the rate of approach determined by k.

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