Question

Plot a direction field and some integral curves for the exact equation $$ \left(3 x^{2}+2 y\right) d x+(2 y+2 x) d y=0 $$ on the rectangle \(\\{-2 \leq x \leq 2,-2 \leq y \leq 2\\} .\) (See Exercise \(37(\mathrm{~b})\) ).

Short answer

Question: Determine the potential function for the exact differential equation \((3x^2 + 2y)dx + (2y + 2x)dy = 0\) and plot its direction field and some integral curves in the rectangle \(-2 \leq x \leq 2,-2 \leq y \leq 2\). Solution: The potential function for the given exact differential equation is \(\phi(x,y) = x^3 + 2xy + y^2 = C\). The direction field and integral curves within the rectangle \(-2 \leq x \leq 2,-2 \leq y \leq 2\) can be plotted using small line segments with the obtained slopes for the first-order DE, and the integral curves will be level curves of the potential function.

Step-by-step solution

Determine if the given equation is exact

The equation takes the form \(M(x, y) dx + N(x, y) dy = 0\). Check if it's exact by calculating \(\frac{\partial M}{\partial y}\) and \(\frac{\partial N}{\partial x}\): $$ M = 3x^2 + 2y \\ N = 2y + 2x \\ \frac{\partial M}{\partial y} = 2 \\ \frac{\partial N}{\partial x} = 2 $$ Since \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\), the given equation is indeed exact.

Find the solution to the exact equation

Integrate \(M\) w.r.t \(x\) and \(N\) w.r.t \(y\) to obtain the potential function $\phi(x,y): $$ \int M dx = \int (3x^2+2y) dx = x^3 + 2xy + f(y) \\ \int N dy = \int (2y+2x) dy = xy + y^2 + g(x) $$ Comparing the two expressions, we see that \(\phi(x,y) = x^3 + 2xy + y^2 = C\), where \(C\) is the constant of integration.

Find the direction field and integral curves within the rectangle

Rewrite the exact equation as the first-order DE \(\frac{dy}{dx} = -\frac{M}{N} = -\frac{3x^2+2y}{2y+2x}\). Then, calculate \(\frac{dy}{dx}\) for a set of points that lie inside the given rectangle, \(-2 \leq x \leq 2,-2 \leq y \leq 2\). Typically, the direction field is depicted using small line segments whose slopes correspond to the values of \(\frac{dy}{dx}\).

Plot the direction field and some integral curves

Create a diagram showing the direction field for the given DE and some integral curves within the rectangle. Plot them on points where \(-2 \le x \le 2\) and \(-2 \le y \le 2\). The integral curves will be the level curves of the potential function \(\phi(x, y) = x^3 + 2xy + y^2 = C\). In conclusion, we have found the solution of the given exact equation, generated the direction field, and plotted several integral curves within the given rectangle.

Key concepts

Direction Field

A direction field is a visual tool used to understand differential equations.
It helps us predict the behavior of solutions without solving them analytically.
Essentially, it consists of small arrows or line segments representing the slope of the solution at various points in a plane.

• The slope at each point is determined by the differential equation itself, specifically the expression \(\frac{dy}{dx}\).
• Each segment's slope gives us a brief overview of how the solution might evolve at that specific point.

Consider our example equation.
We rewrite it as \(\frac{dy}{dx} = -\frac{3x^2+2y}{2y+2x}\), representing the slope of the segments.
By calculating this for points in the rectangle \(-2 \leq x \leq 2\) and \(-2 \leq y \leq 2\), we create a direction field. This field might look complex, but it’s just a map showing possible paths that solutions can take.

Integral Curves

Integral curves are paths that solutions to differential equations follow.
If you imagine each direction field line segment as a moment in time, the integral curves are the full journey, connecting these moments.

• They are solutions to the differential equation represented in the plane, passing through a field of direction lines.
• For the given equation, the integral curves follow the equation \(\phi(x, y) = x^3 + 2xy + y^2 = C\).

Each curve corresponds to a different value of \(C\), a constant representing a particular solution’s level of the potential function.
Imagine these curves as topographical lines on a map, each at a different elevation or energy level.
They smoothly navigate through our direction field, laying out the full path a solution can travel, dictated by its initial conditions.

Potential Function

The potential function, also referred to as \(\phi(x, y)\), consolidates a differential equation into a single expression.
It plays a similar role to potential energy in physics, holding all the information about how a system behaves.

• The potential function can often simplify solving exact differential equations.
• In our solved example, \(\phi(x, y) = x^3 + 2xy + y^2 = C\) is the potential function.

This function emerges from integrating both parts of the exact equation.
It allows us to express the implicit solution in a compact and elegant form.
Solutions to the original equation are represented as level curves (integral curves) of this potential function.
Visualize it as the underlying landscape through which our differential equation's paths, or solutions, travel.

First-order Differential Equations

First-order differential equations involve derivatives of the first order.
They are widely utilized in modeling real-world phenomena, such as population dynamics, heat transfer, and more.

• They consist of equations where a function and its first derivative summarize a system's rate of change.
• These equations can often be represented in the form \(M(x, y) dx + N(x, y) dy = 0\), like our exact differential equation example.

In exact differential equations, solutions are found by utilizing the condition of exactness: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\).
This simplifies solving since it allows integrating directly to find the potential function.
Once the potential function is known, the solutions, effectively the integral curves, are described by it.
First-order differential equations are fundamentally crucial in fields spanning physics, engineering, finance, and beyond.