Problem 12 - Thomas Calculus

Question

Find the family of solutions of the given differential equation and the family of orthogonal trajectories. Sketch both families. $$ \begin{array} { l } { \text { a. } x d x + y d y = 0 } \\ { \text { b. } x d y - 2 y d x = 0 } \end{array} $$

Step-by-Step Solution

Answer

(a) Family of circles: \( x^2 + y^2 = C \), orthogonal lines: \( y = Cx \). (b) Family of parabolas: \( y = Cx^2 \), orthogonal hyperbolas: \( y^2 = -\frac{x^2}{2} + C \).

Step 1: Identify the Type of Differential Equation (Part a)

The differential equation given is \( x\,dx + y\,dy = 0 \). This is a separable differential equation.

Step 2: Separate Variables and Integrate (Part a)

Rewrite the equation as \( x\,dx = -y\,dy \). Integrate both sides: \( \int x\,dx = \int -y\,dy \), which gives \( \frac{x^2}{2} = -\frac{y^2}{2} + C \).

Step 3: Simplify the Solution (Part a)

Simplify the integrated form to get \( x^2 + y^2 = C \), which represents a family of circles centered at the origin.

Step 4: Find the Orthogonal Trajectories (Part a)

To find the orthogonal trajectories, first find the slope \( \frac{dy}{dx} \). For \( x dx + y dy = 0 \), rewrite as \( \frac{dy}{dx} = -\frac{x}{y} \). The orthogonal slope is \( \frac{dy}{dx} = \frac{y}{x} \).

Step 5: Integrate for Orthogonal Trajectories (Part a)

Integrate \( \frac{dy}{dx} = \frac{y}{x} \), which gives \( \int \frac{1}{y} dy = \int \frac{1}{x} dx \). Solving, we get \( \ln |y| = \ln |x| + C \), or \( y = Cx \).

Step 6: Identify the Type of Differential Equation (Part b)

The differential equation given is \( x\,dy - 2y\,dx = 0 \). This is also a separable differential equation.

Step 7: Separate Variables and Integrate (Part b)

Rewrite the equation as \( 2y\,dx = x\,dy \). Separating variables, we get \( \frac{2y}{x} = \frac{dy}{dx} \). Integrate \( \int \frac{1}{y} dy = \int \frac{2}{x} dx \), leading to \( \ln |y| = 2\ln |x| + C \).

Step 8: Simplify the Solution (Part b)

Simplify the equation to get \( y = Cx^2 \), which represents a family of parabolas.

Step 9: Find the Orthogonal Trajectories (Part b)

For the differential equation \( x\,dy - 2y\,dx = 0 \), the slope \( \frac{dy}{dx} = \frac{2y}{x} \). The orthogonal slope is \( \frac{dy}{dx} = -\frac{x}{2y} \).

Step 10: Integrate for Orthogonal Trajectories (Part b)

Integrate \( \frac{dy}{dx} = -\frac{x}{2y} \), yielding \( \int 2y\,dy = -\int x\,dx \). Solving, we obtain \( y^2 = -\frac{x^2}{2} + C \).

Step 11: Sketch Families and Orthogonal Trajectories

For part (a), sketch circles \( x^2 + y^2 = C \) and lines \( y = Cx \). For part (b), sketch parabolas \( y = Cx^2 \) and hyperbolas \( y^2 = -\frac{x^2}{2} + C \). The sketches visually show orthogonality where trajectories intersect.

Key Concepts

Separable Differential Equations

Separable differential equations are equations where variables can be separated on opposite sides of the equation before integration. They take the form of \( f(y)dy = g(x)dx \). This structure allows us to integrate each side independently, making them one of the simpler types to solve:

• **Part (a):** For the equation \( x dx + y dy = 0 \), we rearrange it to \( x dx = -y dy \), clearly separating the variables.
• Integrating both sides leads to solutions like \( \frac{x^2}{2} = -\frac{y^2}{2} + C \), eventually simplified to \( x^2 + y^2 = C \). This represents circles, showing the geometric aspect of differential equations.
• **Part (b):** Similarly, for \( x dy - 2y dx = 0 \), rearranging gives \( \frac{dy}{dx} = \frac{2y}{x} \). Integration yields \( \ln |y| = 2 \ln |x| + C \), simplified to parabolas expressed as \( y = Cx^2 \).

Separable differential equations bridge algebra with geometry, offering solutions like circles and parabolas, highlighting how differential equations describe curves and shapes.

Orthogonal Trajectories

Orthogonal trajectories are curves that intersect a family of curves at right angles. To find them:

• Calculate the slope of the original curve family from the differential equation.
• Adjust to the orthogonal slope, often being the negative reciprocal of the original slope.

For **Part (a):** We determined the slope \( \frac{dy}{dx} = -\frac{x}{y} \) for \( x dx + y dy = 0 \). The orthogonal trajectory has a slope of \( \frac{y}{x} \), leading to linear solutions like \( y = Cx \).
In **Part (b):** The slope for \( x dy - 2y dx = 0 \) was found to be \( \frac{dy}{dx} = \frac{2y}{x} \), with the orthogonal trajectory having a slope of \( -\frac{x}{2y} \). Integration delivers hyperbolic curves such as \( y^2 = -\frac{x^2}{2} + C \). These steps illustrate a neat mathematical dance: altering slopes multiplies the diversity of geometric structure and interactions among curves.

Integral Calculus

Integral calculus is the branch of mathematics focused on integration, the reverse process of differentiation. It enables finding functions represented in terms of areas or cumulative sums. Central to solving differential equations, integration provides exact solutions like:

• Transforming a simple equation into a full-fledged function, giving clear insight into the behavior of systems described by differential equations.
• In our tasks, splitting variables and integrating gives rise to curves such as circles \( x^2 + y^2 = C \) and parabolas \( y = Cx^2 \). These describe foundational geometric shapes.
• Integration of orthogonal trajectory slopes unfolds new structures, adding layers of comprehension to mathematical models.

Embracing integral calculus is embracing the language of change and accumulation, crucial in both mathematical formulations and real-world interpretations, broadening analytical capabilities.