Question

Solve the given differential equation. $$ y^{\prime}=\left(3 x^{2}-1\right) /(3+2 y) $$

Short answer

Question: Solve the differential equation \(y^{\prime} = \frac{3x^2 - 1}{3 + 2y}\). Answer: The solution to the differential equation is given in implicit form as \(3y + y^2 = x^3 - x + C\), where \(C\) is the integration constant.

Step-by-step solution

Rewrite the Differential Equation as a Separable Equation

To rewrite the given equation as a separable differential equation, we can write it in the form \(y^{\prime} = f(x)g(y)\): $$ y^{\prime} = \left(3 x^{2}-1\right) \frac{1}{3+2 y} $$

Separate the Variables

In order to separate the variables, we need to write the equation in terms of \(dy\) and \(dx\). We can do this by multiplying both sides by \((3+2y)dx\): $$ (3+2y)dy = \left(3x^2 - 1\right)dx $$

Integrate Both Sides

Now, we need to integrate both sides of the equation with respect to their variables: $$ \int (3+2y)dy = \int (3x^2 - 1)dx $$

Calculate the Integrals

Calculate the integrals on both sides of the equation: $$ \int (3+2y)dy = 3\int dy + 2\int ydy = 3y + y^2 + C_1 $$ $$ \int (3x^2 - 1)dx = 3\int x^2 dx - \int dx = x^3 - x + C_2 $$

Solve for y

Now that we have integrated both sides, we need to solve for y. We can first combine the integration constants \(C_1\) and \(C_2\) into one constant, say \(C\): $$ 3y + y^2 = x^3 - x + C $$ However, this equation is not easily solvable for \(y\) since it is a quadratic equation. We can leave the solution in implicit form, but the final answer should include the integration constant, \(C\).

Final Solution

The solution to the given differential equation is: $$ 3y + y^2 = x^3 - x + C $$

Key concepts

Differential Equations

A differential equation is an equation that relates a function with its derivatives. In this problem, we were given a differential equation in the form of

• \( y' = \frac{3x^2 - 1}{3 + 2y} \).

This expresses how the derivative of \( y \) is related to \( x \) and \( y \) themselves through a fraction. Differential equations are powerful tools for modeling real-world phenomena where change continuously occurs, such as population growth, heat distribution, or motion.
Solving a differential equation means finding the function, or set of functions, that satisfy the relationship described by the equation. In this case, we are seeking functions \( y(x) \) that meet the equation's requirements. Solution methods can vary, and our method of choice here is "separable" differential equations.

Integration

Integration is a fundamental concept in calculus that involves finding the antiderivative of functions. It's the process of determining the area under curves, amongst other things. When we separate a differential equation into terms involving only \( dx \) and \( dy \), we often need integration to find the function that satisfies the equation.
For this exercise:

• We perform \( \int (3 + 2y) dy = 3y + y^2 + C_1 \),
• And \( \int (3x^2 - 1) dx = x^3 - x + C_2 \).

In integration:

• Each term of the polynomial is integrated separately,
• And then the constants of integration \( C_1 \) and \( C_2 \) are incorporated.

Integrals can be definite, providing a specific numerical value, or indefinite, providing a family of solutions, represented by the constant \( C \). In this case, we deal with indefinite integrals because we have no given bounds.

Implicit Solution

An implicit solution to a differential equation is a solution where the dependent variable is not isolated on one side of the equation. In other words, the relationship between \( y \) and \( x \) is given in a combined form such as

• \( 3y + y^2 = x^3 - x + C \).

Instead of solving explicitly for \( y \) in terms of \( x \), which may not always be possible or easy, we leave the equation in this form. This is common in differential equations, particularly when dealing with non-linear equations where factoring or applying algebraic manipulations to isolate \( y \) directly may be complex.
Such implicit solutions still validly represent the set of functions that satisfy the original differential constraint even if they aren't neatly expressed as \( y = f(x) \). Application or interpretation of these implicit solutions can follow similar processes as explicit solutions, often by analyzing the equation's behavior or using numerical methods.

Variable Separation

Variable separation is a technique used to solve differential equations by rearranging the equation such that all terms involving one variable are on one side and all terms involving the other variable are on the other.
In our problem, we started with the transformed equation

• \( y' = \frac{3x^2 - 1}{3 + 2y} \),

and rewrote it as a separable form:

• \( (3 + 2y) dy = (3x^2 - 1) dx \).

This step is crucial, as it allows us to integrate each side independently:

• The left side in terms of \( y \),
• And the right side in terms of \( x \),

to find a relationship between \( x \) and \( y \).
Separable differential equations are among the simplest types of differential equations to solve, as this method leverages basic integration skills for solutions. Once separated, solving involves straightforward integration and the inclusion of a constant of integration to represent all potential solutions.