Geometric Interpretation of ODEs 2

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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.

Sketch and Solve $\frac{dy}{dx} = x + y$.

Let’s analyze and solve this first-order ordinary differential equation (ODE). We will start with a qualitative approach by sketching the direction field and then proceed to solve the equation analytically.

A direction field (or slope field) is a graphical tool that shows the slope of the solution curve $\frac{dy}{dx}$ at various points (x, y). This helps us visualize how solutions behave without explicitly solving the equation.

An isocline is a curve along which the slope $\frac{dy}{dx}$ remains constant. For our differential equation, $\frac{dy}{dx} = x + y = C$ where C is a constant, each value of C corresponds to a different isocline.

Solving for y, we get y = -x + C. Thus, each isocline is a straight line of the form y = -x + C with a slope of −1. Each line corresponds to a different value of C and represents a set of points where the slope $\frac{dy}{dx}$ is constant

Specific Isoclines: Let’s consider specific values of C to understand how the direction field behaves along different isoclines.

• For C = 0, we have the equation of the isocline y = -x, meaning that along this line, the direction field has a constant slope of y’ = 0, i.e., the line segments in the direction field (solution curves near this line) are horizontal along this line.
• For C = -1, the isocline is y = -x - 1, meaning that on this line the direction field has a constant slope -1, indicating line segments inclined downward (slanting) at a 45-degree angle.
• For C = 1, the isocline is y = -x + 1. The slope of the direction field along this line is one, so the line segments are inclined upward a 45-degree angle.
• For C = 2, the isocline is y = -x +2, the slope y’ = 2, indicating steeper upward line segments.
• For C = -2, we get the line y = -x -2, meaning that on this line the direction field has a constant slope -2, representing steeper downward line segments (slopes) and so on.

Observing Solution Behavior with Isoclines:

1. Near the line y = -x, solution curves are nearly horizontal or have minimal slopes.
2. Above the line y = -x, the sum (slope $\frac{dy}{dx}$) x + y is positive, and the solution curves are inclined upwards. These solution curves more steeply the further they are from the line y = −x (as x→∞, x→-∞). These solution curves approaches the line y = -x +1 asymptotically, without touching it. This means that as x increases or decreases without bound, the solutions get closer to y = −x +1 but remain distinct.
3. Below the line, y = -x, the sum x + y is negative, hence the slope is negative and the solution curves are incline downwards. As x → ∞, the solution curves continue to fall more steeply (without bond) the further away they are from the line y = −x.

Solving the Differential Equation Analytically

To deepen our understanding, let’s solve the differential equation analytically: $\frac{dy}{dx} = x + y ↭ \frac{dy}{dx}-y = x$. This is a linear first-order differential equation of the form: $\frac{dy}{dx} + P(x)y = Q(x)$ where P(x) = -1 and Q(x) = x.

1. Compute the integral factor: $μ(x) = e^{\int P(x)dx} = e^{\int -dx} = e^{-x}$

2. Multiply both sides of the ODE by μ(x): $e^{-x}\frac{dy}{dx}-e^{-x}y = e^{-x}x$

3. Recognize the Left Side as a Product Derivative: $\frac{d}{dx}(e^{-x}y) = e^{-x}x$

4. Integrate both sides with respect to x: $e^{-x}y = \int e^{-x}xdx$

   Use integration by parts u = x ⇒ du = dx, dv = e^-xdx ⇒ v = -e^-x

   $e^{-x}y = \int e^{-x}xdx = -xe^{-x}-\int -e^{-x}dx ⇒ e^{-x}y = -xe^{-x} - e^{-x} + C$

5. Multiply both sides by e^x: y = -x - 1 + Ce^x. The general solution to the differential equation is: y(x) = -x - 1 + Ce^x

• As x → -∞, e^x → 0, y(x) ≈ -x - 1. The solution curves approach the line y = -x - 1. This means that as x goes to −∞, the solutions approach the line y = −x -1 asymptotically.
• As x → ∞, e^x → ∞, the term Ce^x dominates. If C > 0, y(x)→ ∞. If C < 0, y(x)→ -∞ (Refer to Figure i for a visual representation and aid in understanding it).

Sketch and Solve $y’ = \frac{-y}{x^2+y^2}$

This is a nonlinear ordinary differential equation (ODE) due to the presence of y² in the denominator. Solving this equation explicitly for y(x) in terms of elementary functions is quite challenging. However, we can gain significant insights by analyzing the isoclines and the behavior of the solutions.

An isocline is a curve along which the slope y’ of the solution curves is constant. That is, all solution curves intersecting an isocline have the same slope at the points of intersection. We set y′ = m (a constant) and analyze the resulting equation.

$\frac{-y}{x^2+y^2} = m$

1. Case 1. If m = 0 ⇒ y = 0. The x-axis (y = 0) is an isocline where the slope y’ = 0, meaning the line segments in the direction field are horizontal along this line.
2. Case 2. m ≠ 0, $\frac{-y}{x^2+y^2} = m ↭[\text{Multiplying both sides by x² + y²}] -y = m(x^2+y^2) ↭\frac{-1}{m}y = x^2 + y^2 ↭ x^2 + y^2 + \frac{1}{m}y = 0 ↭[\text{Complete the square}] x^2 + (y+\frac{1}{2m})^2 = \frac{1}{4m^2}$. For each constant slope m ≠ 0, the isoclines is a circle centered along the y-axis at (0, $\frac{-1}{2m}$) with radius r = $\frac{1}{2|m|}$. The sign of m determines the position of the center along the y-axis.

Examples of Isoclines. Let’s consider specific values of m to illustrate the isoclines:

• m = -1 ⇒ We have a circle centered at (0, $\frac{1}{2}$) and radius $\frac{1}{2}$, and the line segments in the direction field are inclined downwards at a 45-degree angle (a slope y’ = 1).
• m = 1 ⇒ We have a circle centered at (0, $\frac{-1}{2}$) and radius $\frac{1}{2}$, and the line segments in the direction field are inclined upwards at a 45-degree angle.
• m = -2 ⇒ We have a circle centered at (0, $\frac{1}{4}$) and radius $\frac{1}{4}$, and the line segments in the direction field are inclined steeper downward compared to m = -1.
• m = 2 ⇒ We have a circle centered at (0, $\frac{-1}{4}$) and radius $\frac{1}{4}$, and the line segments in the direction field are inclined steeper upward compared to m = 1.

Behavior of Solutions

1. y = 0 is a constant solution of our EDO. The Existence and Uniqueness Theorem ensures that solutions starting on one side of the x-axis remain on that side, provided the right-hand side of the ODE is continuous.
2. Solutions Above the x-axis (y > 0). The numerator -y is negative, and the denominator x² + y² is always positive. Thus, y’ < 0 (the slopes are negative). As x increase, x² dominates y², and the magnitude of y’ decreases (x² + y² grows larger), meaning solutions approach y = 0 asymptotically from above but never reach it (Refer to Figure ii for a visual representation and aid in understanding it).
3. Solutions Below the x-axis (y < 0). The numerator -y is positive and the denominator is always positive, therefore y’ > 0. As x increase, similar to above, x² dominates y², and the magnitude of y’ decreases (x² + y² grows larger). Solutions approach y = 0 asymptotically from below but never reach it.
4. In general, all non-zero solutions flatten out (approach the x-axis) as x→∞, but never touch it. Solutions flatten out as they extend along the x-axis, reflecting the diminishing influence of y in the denominator compared to x².

1. Above the x-axis: Negative slopes decreasing in magnitude with increasing x.
2. Below the x-axis: Positive slopes decreasing in magnitude with increasing x.

Solve and sketch y’ = y -2x with initial condition (1, 0)

Solve the Differential Equation

This is a linear first-order differential equation. We can solve it using an integrating factor.

1. Rewrite the Equation in Standard Linear Form y’ + P(x)y = Q(x): y’ - y = -2x
2. Integrating Factor: μ(x) = $e^{\int -1dx}=e^{-x}$.
3. Multiply Both Sides by Integrating Factor: $e^{-x}y’ -e^{-x}y = -2xe^{-x} ↭ \frac{d}{dx}(e^{-x}·y) = -2xe^{-x}$
4. Integrate Both Sides: $\int \frac{d}{dx}(e^{-x}·y)dx = \int -2xe^{-x}dx ↭ e^{-x}·y = \int -2xe^{-x}dx ↭ e^{-x}·y = (2xe^{-x} + 2e^{-x}) + C$.
5. Solve for y(x): $y = 2x + 2 + Ce^{x}$
6. Apply Initial Condition y(1) = 0: $0 = 2·1 + 2 + C·e^{1} ⇒ C = -4e^{-1}$
7. Particular Solution Passing Through (1, 0): $y(x) = 2x + 2 + -4e^{x-1}$

Sketch the Direction Field

A direction field provides a visual representation of the slopes of the solution curves at various points in the plane.

Compute Slopes at Selected Points. Choose a grid of points (x, y) and compute the slope y′= y −2x at each point.

Along the line y = 2x, the slope y′ = 0. This means solution curves have horizontal tangents when y = 2x.

Plotting the particular solution. Start at (1, 0). As x increases, y(x) decreases rapidly due to the exponential term (pulling y downward exponentially). As x decreases, the exponential term diminishes, and y(x) approaches the line y = 2x +2 (Refer to Figure i for a visual representation and aid in understanding it).

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