Problem 36 - Thomas Calculus

Question

a. Use a CAS to plot the slope field of the differential equation $$y^{\prime}=\frac{3 x^{2}+4 x+2}{2(y-1)}$$ over the region \(-3 \leq x \leq 3\) and \(-3 \leq y \leq 3\) b. Separate the variables and use a CAS integrator to find the general solution in implicit form. c. Using a CAS implicit function grapher, plot solution curves for the arbitrary constant values \(C=-6,-4,-2,0,2,4,6\) d. Find and graph the solution that satisfies the initial condition \(y(0)=-1\)

Step-by-Step Solution

Answer

Plot the slope field, separate and integrate variables; find solutions for constants and initial conditions.

Step 1: Plot the Slope Field

Utilize a computer algebra system (CAS) like Desmos or GeoGebra to input the differential equation \(y' = \frac{3x^2 + 4x + 2}{2(y-1)}\). Set the window to display the region where \(-3 \leq x \leq 3\) and \(-3 \leq y \leq 3\). Plot the slope field by visualizing numerous small line segments, where the slope of each segment is given by the differential equation at that point.

Step 2: Separate the Variables

Rewrite the differential equation \(y' = \frac{3x^2 + 4x + 2}{2(y-1)}\) as \((y-1) dy = \left(\frac{3x^2 + 4x + 2}{2}\right) dx\). This separates the variables, placing all \(y\) terms on one side and all \(x\) terms on the other side of the equation.

Step 3: Integrate Both Sides

Integrate both sides separately. The left-hand side becomes \(\int (y-1) \, dy = \frac{1}{2}y^2 - y + C_1\) and the right-hand side becomes \(\int \frac{3x^2 + 4x + 2}{2} \, dx\). Solve the right-hand side to obtain \(\frac{1}{2}x^3 + x^2 + x + C_2\). Combine the resulting expressions to find the implicit general solution.

Step 4: Express the General Solution

Combine the results from the integration step to obtain the general implicit solution: \[ \frac{1}{2}y^2 - y + C_1 = \frac{1}{2}x^3 + x^2 + x + C_2 \]. This expression represents the general solution of the differential equation.

Step 5: Plot Solution Curves for Specific Constants

Use a CAS implicit function grapher to plot the general solution \(\frac{1}{2}y^2 - y = \frac{1}{2}x^3 + x^2 + x + C\) for various values of \(C\): \(-6, -4, -2, 0, 2, 4, 6\). Each curve represents a solution to the differential equation for a particular constant.

Step 6: Solve with Initial Condition

Substitute the initial condition \(y(0) = -1\) into the general solution. This gives \(\frac{1}{2}(-1)^2 + 1 = C\), simplifying to \(\frac{1}{2} + 1 = C\), and thus \(C = \frac{3}{2}\). Insert \(C = \frac{3}{2}\) back into the implicit solution to represent the specific curve satisfying the initial condition.

Step 7: Graph the Particular Solution

Use a CAS to graph the particular solution obtained from Step 6: \(\frac{1}{2}y^2 - y = \frac{1}{2}x^3 + x^2 + x + \frac{3}{2}\). This curve will pass through the point \((0, -1)\) and represents the specific solution that satisfies the given initial condition.

Key Concepts

Slope Field

A slope field is a graphical representation of a differential equation. It consists of small line segments or arrows that indicate the slope or direction of the solution at any given point. In this exercise, the slope field represents the differential equation \( y' = \frac{3x^2 + 4x + 2}{2(y-1)} \) over the region \(-3 \leq x \leq 3\) and \(-3 \leq y \leq 3\). By plotting these segments, it provides a visual insight into the behavior and potential paths of solutions. Using a Computer Algebra System (CAS), like Desmos or GeoGebra, makes it easy to generate and analyze such fields by automatically computing and displaying the slopes at multiple points. Slope fields are especially helpful for understanding complex differential equations for which analytic solutions might not be straightforwardly derived.

Variable Separation

Variable separation is a common technique used in solving differential equations. The idea is to rearrange the equation so that all terms involving the dependent variable \(y\) are on one side, while terms involving the independent variable \(x\) are on the opposite side. This allows each variable to be integrated separately.
For the given differential equation \( y' = \frac{3x^2 + 4x + 2}{2(y-1)} \), we can rewrite it in the form \( (y-1) dy = \left(\frac{3x^2 + 4x + 2}{2}\right) dx \). This separation facilitates the integration of both sides, subsequently leading to the solution of the equation. It simplifies the process by breaking down the problem into more manageable integrals that can be solved to find the implicit solution.

CAS Integrator

A CAS integrator is a powerful tool that performs symbolic computations, including integration, which is a critical step in solving differential equations. By utilizing a CAS, you can quickly and accurately integrate both sides of a separated variable differential equation, like in our exercise. This step transforms the differential equation into an implicit solution that combines the integrals from both sides.

• The left side integrates to \( \int (y-1) \, dy = \frac{1}{2}y^2 - y + C_1 \).
• The right side integrates to \( \int \frac{3x^2 + 4x + 2}{2} \, dx = \frac{1}{2}x^3 + x^2 + x + C_2 \).

The CAS assists in handling complex algebraic manipulations, ensuring a precise outcome with much less effort than manual calculations. It's invaluable for students and professionals tackling intricate mathematical problems.

Particular Solution

The particular solution involves finding a specific solution that satisfies given initial conditions or particular constraints. In this exercise, the initial condition \( y(0) = -1 \) is used to determine a specific value of the arbitrary constant \(C\) in the general solution.
By substituting \( x = 0 \) and \( y = -1 \) into the implicit solution equation \( \frac{1}{2}y^2 - y = \frac{1}{2}x^3 + x^2 + x + C \) and solving for \(C\), we find:

• \( \frac{1}{2}(-1)^2 - (-1) = C \), which simplifies to \( \frac{1}{2} + 1 = C \).
• Thus, \( C = \frac{3}{2} \).

Inserting \( C = \frac{3}{2} \) back into the equation provides the particular solution
\( \frac{1}{2}y^2 - y = \frac{1}{2}x^3 + x^2 + x + \frac{3}{2} \). This specific equation is then plotted as a curve that passes through the point \((0, -1)\), illustrating a unique path among the general solutions that aligns with the initial condition.