Question

Using Separation of Variables Find a general solution to the differential equation \(y^{\prime}=\left(x^{2}-4\right)(3 y+2)\) using the method of separation of variables.

Short answer

The general solution is \( y = \pm \frac{e^{x^3 - 12x + 3C} - 2}{3} \).

Step-by-step solution

Identify and Rearrange Terms

The given differential equation is \( y' = (x^2 - 4)(3y + 2) \). To apply the method of separation of variables, we need to rearrange terms such that all \( y \) terms are on one side and all \( x \) terms are on the other. Divide both sides by \( (3y + 2) \) and multiply both sides by \( dx \):\[ \frac{1}{3y + 2} dy = (x^2 - 4) dx. \]

Integrate Both Sides

Integrate both sides of the equation to find the general solution. For the left side:\[ \int \frac{1}{3y + 2} dy = \frac{1}{3} \ln|3y + 2| + C_1. \]For the right side:\[ \int (x^2 - 4) dx = \frac{x^3}{3} - 4x + C_2. \]

Combine and Simplify

Combine the results of the integrations into a single equation:\[ \frac{1}{3} \ln|3y + 2| = \frac{x^3}{3} - 4x + C. \]Here, \( C = C_2 - C_1 \) is a constant representing the combination of both integration constants.

Solve for \( y \)

To solve for \( y \), first clear the fraction by multiplying through by 3:\[ \ln|3y + 2| = x^3 - 12x + 3C. \]Exponentiate both sides to eliminate the natural logarithm:\[ |3y + 2| = e^{x^3 - 12x + 3C}. \]So, solving for \( y \):\[ 3y + 2 = \,\pm e^{x^3 - 12x + 3C} \]\[ y = \,\pm \frac{e^{x^3 - 12x + 3C} - 2}{3}. \]

Key concepts

Separation of Variables

Separation of variables is a powerful technique often used to solve differential equations. This method involves rearranging a differential equation so that each variable is isolated on opposite sides. The goal is to express the equation in the form of two integrals, one in terms of each variable. This helps simplify the process of solving the equation.

In this exercise, the differential equation given is:\[ y' = (x^2 - 4)(3y + 2) \]To separate variables, we manipulate the equation by dividing both sides by \(3y + 2\) and multiplying by \(dx\). This allows the equation to be expressed as:\[ \frac{1}{3y + 2} dy = (x^2 - 4) dx \]By isolating \(dy\) and \(dx\) on separate sides of the equation, we can then integrate both expressions independently. This organization is the key to applying the technique effectively.

Integration

Once we've separated the variables, we move on to integrating both sides of the equation. Integration is the process of finding the integral, or the antiderivative, of a function. This process is essential in solving differential equations as it helps to transition from the rate of change (the derivative) back to the original function.

• The left side of the equation involves integrating \(\frac{1}{3y + 2} dy\). The solution to this integration is \(\frac{1}{3} \ln|3y + 2|\).
• On the right side, the integration of \((x^2 - 4) dx\) results in \(\frac{x^3}{3} - 4x\).

These integration steps convert both sides into expressions that describe how \(y\) and \(x\) relate to one another, which are crucial for finding the general solution. It is important to remember to include the integration constants, represented as \(C_1\) and \(C_2\), because integration is an indefinite process.

General Solution

The culmination of separating variables and integrating is to express the general solution, which describes the function explicitly. After performing the integration, we combine both expressions into one equation:\[ \frac{1}{3} \ln|3y + 2| = \frac{x^3}{3} - 4x + C \]Here, \(C\) represents a constant that results from the subtraction of the integration constants \(C_1\) and \(C_2\).

• To solve for \(y\), the next step involves removing the natural logarithm by exponentiating both sides.
• This leads to solving for \(y\), yielding: \(y = \pm \frac{e^{x^3 - 12x + 3C} - 2}{3}\).

The general solution accounts for the constant \(C\), which generalizes the family of solutions that satisfy the original differential equation. This constant represents the possible vertical shifts in the solution, showing how integration introduces a family of curves rather than a single curve.