Geometric Interpretation of ODEs

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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, $\frac{dy}{dx}$

Solving differential equations

To solve ordinary differential equations (ODEs), several methods can be employed, each suited to different types of equations. Here are some of the key methods:

• Separation of Variables. This method is used for first-order ODEs that can be written in the form $\frac{dy}{dx} = g(x)h(y)$. By separating the variables, you integrate both sides separately: $\int \frac{dy}{h(y)} = \int g(x)dx + C$

• Integrating Factors. This method is applied to linear first-order ODEs of the form $\frac{dy}{dx} + P(x)y = Q(x)$. The idea is to multiply the entire equation by an integrating factor μ(x) = $e^{\int P(x)dx}$, which simplifies the left-hand side into a product derivative, making the equation easier to solve: $\frac{d}{dx}(y·μ(x)) = Q(x)·μ(x)$. Integrating both sides gives the solution.

• Exact Differential Equations. These are ODEs that can be expressed in the form of a total differential of a function, M(x,y)dx + N(x,y)dy = 0, where the exactness condition $\frac{∂M}{∂y} = \frac{∂N}{∂x}$ must be satisfied. If the equation is exact, you should find the potential function f(x, y) such that $\frac{∂f}{∂x} = M$ and $\frac{∂f}{∂y} = N$. The solution is f(x,y) = C.

• Geometric Interpretations. This approach involves visualizing the solution in the phase plane or using graphical methods to understand the behavior of the solutions. For example, in autonomous differential equations, the phase plane can help in understanding the equilibrium points and the stability of the solutions.

• Numerical Methods. For ODEs that cannot be solved analytically, numerical methods are used to find approximate solutions.

Introduction to First-Order Ordinary Differential Equations (ODEs)

Our primary goal is to understand how to solve first-order ODEs analytically. These equations take the form: $y’ = f(x, y)$ (1).

This equation indicates how the derivative of a function y depends on both the independent variable x and the dependent variable y. In simple terms, y′(x) gives the rate of change of y with respect to x, and f(x, y) describes how this rate changes depending on the values of x and y.

Examples of First-Order ODEs

Here are some common examples of first-order ODEs: $y’ = \frac{x}{y}, y’ = x -y^2, y’ = y -x^2.$ These are first-order ODEs because they involve the first derivative y′(x) of the unknown function y(x), and no higher derivatives. This characteristic makes them first-order ODEs.

To uniquely determine a solution, we often use an initial condition: y(x₀) = y₀, x₀, y₀ ∈ ℝ (2). This condition specifies the value of the solution y(x) at a particular point x₀. Solving an ODE typically means finding a function y(x) that satisfies both the differential equation (1) and the initial condition (2).

Geometric Interpretation of ODEs

Before we dive into analytical solutions methods, it’s useful to explore the geometric interpretation of the ODE in equation (1). This will help us visualize the behavior of solutions to differential equations, which is an essential tool in understanding their properties.

What is a Solution to an ODE?

A function y₁(x) is a solution to the ODE (1) with initial condition (2) over an interval I = (a, b) if:

• It satisfies the initial condition: y₁(x₀) = y₀.
• It has a continuous first derivative on the interval I, meaning y₁ ∈ C⁽¹⁾(I; ℝ).
• It satisfies the equation (1): y₁’(x) = f(x, y₁(x)) for all x ∈ I.

This means that the solution y₁(x) gives a specific relationship between x and y, where the rate of change of y at any point x matches the function f(x, y₁(x)).

Integral Curves

Definition. The graph of a solution y = y₁(x) to the ODE (1) with the initial condition (2) is called an integral curve of equation.

At any point (x₀, y₁(x₀)) on the curve, the derivative $y’_1(x_0)$ represents the slope of the tangent line to the curve at that point. Geometrically, this slope is the tangent of the angle at which the tangent line intersects the x-axis. From equation (1), we know that the slope at any point (x₀, y₁(x₀)) is simply given by f(x₀, y₁(x₀)).

If we change the coordinates (x₀, y₁(x₀)) to some other point (x₁, y₁(x₁)), the slope at that new point will be f(x₁, y₁(x₁)). By calculating the slope at various points, we can draw these slopes as small line segments attached to each point in the plane. When we do this for many points, these segments collectively form what is known as the direction field (or slope field) of the ODE.

The direction field is a graphical representation of the behavior of solutions to the ODE. It consists of small line segments, where each segment represents the slope of the solution curve at that point, as determined by the function f(x, y).

• Each line segment at a point (x, y) has a slope equal to f(x, y).
• By visualizing these segments, we can sketch the overall shape of solution curves, even without solving the ODE explicitly.

Now consider a curve that is tangent to the given direction field at each point (meaning this curve is tangent to the line segments at each point). Such a curve is an integral curve of the ODE. Indeed, at each point, the slope y′(x) of this curve is given by construction as f(x, y), which exactly means that this curve satisfies the differential equation and represents a solution to the ODE.

Isoclines and Their Role

To better understand the direction field, we introduce the concept of isoclines. An isocline is a curve along which the slope of the direction field is constant. Isoclines help us organize the direction field, making it easier to plot and understand.

Isoclines are defined by the equation: f(x, y) = C, where C is a constant. For each fixed value C, we get a curve where the slope of the direction field is constant and equal to C.

Plotting the Direction Field Using Isoclines

To plot the direction field, we follow these steps:

• Fix a constant C and solve the equation f(x, y) = C. This gives a curve where the slope of the direction field is equal to C (constant) at every point.
• Plot this curve in the xy-plane and draw small line segments with slope C at several points along the curve.
• Repeat this process for different values of C, obtaining a family of curves, each representing different constant slopes.
• As you vary C, you can build up a comprehensive picture of the direction field. The general pattern of the direction field becomes clearer as you plot more isoclines with different slopes.

Solve examples

Direction field y’ = ^-x⁄_y

Let’s consider the differential equation y’ = ^-x⁄_y. To find the isoclines for this equation, we set the slope equal to a constant C: ^-x⁄_y = C ⇒[Solving for y, we find] y = ^-x⁄_C.

This equation represents a family of straight lines passing through the origin with various slopes depending on the constant C. Here’s how different values of C affect the direction field:

• For C = 1: The equation becomes y = −x, meaning that along this line, the slope of the direction field is constantly 1 at every point (45°).
• For C = −1: The equation becomes y = x, meaning that along this line, the slope is constantly −1.
• For C = 0, we get $\frac{-x}{y}$ = 0 ⇒ -x = 0 ⇒ x = 0, so the isocline is x = 0, which corresponds to the y-axis. The slope of the direction field along this line is zero, meaning the line segments on the y-axis are horizontal.
• For C = ∞, the slope tends to zero, implying that y → 0, which corresponds to the x-axis. Here, the direction field’s slope tends toward infinity, indicating vertical line segments along the x-axis.

Geometric Interpretation and Integral Curves

The integral fields (i.e., the solution curves) for this differential equation are circles centered at the origin with various radii. Specifically, these curves are defined by equations of the form $x^2+y^2=c_1^2$ where c₁ is a constant that determines the radius of each circle. This family of circles represents the set of all possible solutions to the ODE, depending on different initial conditions.

We can solve the ODEs analytically by separating variables to confirm this family of solutions (Refer to Figure A for a visual representation and aid in understanding it).

The geometric interpretation of first-order ODEs, especially using direction fields and isoclines, provides a valuable tool for visualizing and understanding the behavior of solutions. By plotting the direction field, we can see how the solution curves evolve and get a sense of the qualitative behavior of the ODE, even when an explicit solution is difficult or impossible to find.

Solving the ODE Analytically

Given the differential equation: $y’ = \frac{-x}{y}↭[\text{This can be written as:}] \frac{dy}{dx} = \frac{-x}{y}$

Separation of variables: ydy = -xdx.

Now, we can integrate both sides: $\int ydy = \int -xdx ⇒[\text{This yields:}] \frac{y^2}{2} = \frac{-x^2}{2} + C$ where C is the constant of integration. Multiply both sides by 2 to eliminate the fractions: $y^2 = -x^2 + 2C ↭ x^2 + y^2 = 2C$. This is the equation of a circle centered at the origin with radius $\sqrt{2C}$.

For simplicity, let rename the constant C as $\frac{C_1^2}{2}$, so the equation becomes: $x^2+y^2 = C_1^2$. Thus, the general solution to the ODE is: $y = \sqrt{C_1^2-x^2}$. The complete family of solutions includes both the positive and negative square roots: $y = ±\sqrt{C_1^2-x^2}$

These curves represent circles centered at the origin, as anticipated from the geometric interpretation.

Direction field for y’ = 1 + x - y

Let’s examine the differential equation y’ = 1 + x - y. This means that the slope depends on both x and y, and it can change across the plane.

The direction field for a differential equation is a graphical representation that helps us visualize the behavior of solutions without explicitly solving the ODE. In a direction field:

• Each point (x, y) on the plane, we calculate the slope y’ and draw a small line segment with a slope corresponding to the value of y’ at that point.
• Collectively, these segments illustrate or suggest how solution curves might flow across the plane, (Refer to Figure B for a visual representation and aid in understanding it).

For the equation y′ = 1 +x −y, let’s calculate the slopes y′ at various points and draw small line segments to represent these slopes. This process, if done for many points, will yield a direction field that visually illustrates potential solution curves.

An isocline is a curve along which the slope of the direction field y’ is constant. To find the isoclines for our equation, we set the slope equal to a constant C: y’ = C.

So, the family of isoclines is given by 1 + x -y = C or y = x + (1 - C). This represents a family of straight, parallel lines with slope 1. Each line corresponds to a different value of C and helps us organize and understand the direction field.

Examples of Isoclines. Here are some specific values for C that help illustrate how the direction field behaves along these isoclines:

• For C = 0, we have the line y = x + 1, meaning that on this line, the direction field has a constant slope of y’ = 0, i.e., the line segments along this line are horizontal.
• For C = -1, we get the line y = x + 2, meaning that on this line the direction field has a constant slope -1, indicating line segments slanting downward at a 45° angle.
• For C = 1, this yields y = x. The slope of the direction field along this line is one, so the line segments along this line are inclined upward at a 45° angle.
• For C = 2, we obtain y = x -1, with a slope of y’ = 2, meaning the line segments are even steeper and slanting upward.

Between the isoclines corresponding to C = 0 and C = 2, we observe that the integral curves (solution curves) cannot escape or are confined within a corridor of isoclines unable to cross certain boundaries. All solutions in this region asymptotically approach the line y = x as x → ∞. This line y = x acts as an attractor for solutions in this region.

The line y = x is indeed a solution to the differential equation: y’ = 1 + x -y ↭ 1 = 1 + x -x ∎

Existence and Uniqueness Theorem

The Existence and Uniqueness Theorem provides crucial insight into the behavior of solutions to first-order differential equations ODEs. It tells us when we can expect a unique solution that passes through a given point.

For an equation of the form y’ = f(x, y):

• If the function f(x, y) (the right-hand side of the ODE) in y’ = f(x, y) is continuous in a neighborhood around a point (x₀, y₀) and
• The partial derivative with respect to y, $\frac{∂f}{∂y}$, is also continuous in a neighborhood around (x₀, y₀).

Then the differential equation y' = f(x, y) has a unique solution to the initial value problem passing through the point (x₀, y₀). In other words, two solution curves cannot cross or touch each other in this region, as that would contradict uniqueness.

For our equation:

• f(x, y) = 1 + x -y is continuous everywhere.
• $\frac{∂f}{∂y} = -1$ is also continuous everywhere

Thus, the Existence and Uniqueness Theorem guarantees that for any initial condition y(x₀) = y₀, there exists a unique solution to the ODE passing through the point (x₀, y₀). This also means that two solution (integral) curves cannot cross or touch each other, as this would violate uniqueness.

Analytical Solution: integrating factors

Let’s now find an explicit solution to the ODE by using the method of integrating factors.

The differential equation is: y’ = 1 + x -y ↭[Bring all terms involving y to one side] y’ + y = 1 + x. This is a linear ODE of the form: y’ +p(x)y = q(x) where p(x)= 1 and q(x) = 1 + x.

The integrating factor μ(x) is given by: $μ(x) = e^{\int p(x)dx} = e^{\int 1dx}= e^x$.

Multiplying both sides of the equation by e^x, we get: $e^x·y’ + e^x·y = (1+x)e^x$

Recognize the left-hand as a product derivative of e^x·y: $\frac{d}{dx}(e^x·y) = (1+x)e^x$

Now, integrate both sides with respect to x:

$e^x·y = \int (1+x)e^xdx = \int e^xdx + \int xe^xdx$ 🚀

The first integral is straightforward: $\int e^xdx = e^x$.

The second integral requires integration by parts. Let u = x and dv = e^xdx, so that du = dx and v = e^x.

$\int xe^xdx = xe^x -\int e^xdx = xe^x -e^x$.

Combine results and the full integral becomes: $\int (1+x)e^xdx = e^x + (xe^x -e^x) = xe^x + C$.

Substitute this back into the equation 🚀: $e^x·y = xe^x + C$ where C is the constant of integration. Finally, divide both sides by e^x (which is always positive) to solve for y: y = x + Ce^-x.

As x → ∞, the term Ce^-x tends to zero because e^-x decays exponentially. Therefore, the solution asymptotically approaches the line y = x as x → ∞. This matches our earlier geometric observation that y = x is an attractor.

If an initial solution is given, such as y(x₀) = y₀, we can substitute these values into the general solution to find the specific value of C. For instance, if y(0) = y₀, we have y₀ = 0 + Ce⁰ = C, so C = y₀. Therefore, the solution simplifies to y = x + y₀e^-x.

More specifically, if y₀ = 0, then the constant C equals 0, and the solution simplifies to y = x with slope 1 (y’ = 1), which is also the solution we obtained geometrically.

Solve the Differential Equation xy’ = y -1

Let’s solve the first-order ordinary differential equation (ODE): xy’ = y - 1 ↭[This can be rewritten as] $x·\frac{dy}{dx} = y - 1 ↭ \frac{dy}{dx} =\frac{y-1}{x}$

This is a separable differential equation, meaning we can separate the variables y and x to solve it. Let’s proceed by separating the terms involving y on one side and those involving x on the other side:

$\frac{dy}{y-1} = \frac{dx}{x}$

Now we integrate both sides.

$\int \frac{dy}{y-1} = \int \frac{dx}{x} ↭[\text{This yields}] ln|y-1| = ln|x| + C$ where C is the constant of integration.

Next, we exponentiate both sides to eliminate the logarithms:

$e^{ln|y-1|} = e^{ln|x|+ C}↭ |y-1| = e^C|x|$. Since e^C is always a positive constant, we can replace it with a new constant C (to simplify notation).

At this point, we drop the absolute values, leading to y - 1 = Cx where C is a constant which might absorb the sign difference (both/one of x and/or y-1 are positive or negative y -1 = Cx and/or y -1 = -Cx) and is determined by initial conditions (Refer to Figure C for a visual representation and aid in understanding it).

Thus, the general solution to the differential equation is y(x) = 1 + Cx.This represents a family of straight lines with varying slopes determined by the constant C. Each line passes through the point (0, 1) when x = 0, and the slope of each line is given by C.

Analysis with the Existence and Uniqueness Theorem

The existence and uniqueness theorem for differential equations states that, under certain conditions, a unique solution exists for an initial value problem (an ordinary differential equation together with an initial condition). Specifically, for a first-order equation of the form: y’ = f(x, y) a unique solution exists through the point (x₀, y₀) if:

1. f(x, y) is continuous in a neighborhood of (x₀, y₀).
2. The partial derivative $\frac{∂f}{∂y}$ is continuous in the same neighborhood.

For our equation in its rewritten form: $\frac{dy}{dx} = \frac{y-1}{x}$. We observe that the right-hand side of this equation, $\frac{y-1}{x}$, is not defined when x = 0. This introduces a singularity at x = 0 where the function is obviously not continuous. Therefore,

• For points where x ≠ 0, the function $\frac{y-1}{x}$ is continuous, and the theorem guarantees a unique solution passing through any point (x₀, y₀) where x₀ ≠ 0.
• At points where x = 0, the function $\frac{y-1}{x}$ is not defined, so the existence and uniqueness theorem does not apply at these points.
• The point (0, 1) on the y-axis is special because the differential equation has multiple solutions passing through this point. Specifically, for x = 0, any value of C would satisfy the equation y = 1 + Cx, because the equation degenerates to y = 1 when x = 0. Therefore, we have infinitely many solutions passing through (0, 1), meaning that uniqueness fails at this point.

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