Question

In Problems \(5-14\), first find the general solution (involving a constant \(\mathrm{C}\) ) for the given differential equation. Then find the particular solution that satisfies the indicated condition. $$ \frac{d y}{d x}=x^{2}+1 ; y=1 \text { at } x=1 $$

Short answer

The particular solution is \( y = \frac{x^3}{3} + x - \frac{1}{3} \).

Step-by-step solution

Identify the differential equation and condition

The given differential equation is \( \frac{dy}{dx} = x^2 + 1 \), and we are given the condition \( y = 1 \) at \( x = 1 \).

Find the general solution

To solve the differential equation, integrate both sides with respect to \( x \). \[ \int dy = \int (x^2 + 1) \, dx \]The left side integrates to \( y \) and the right side integrates to \( \frac{x^3}{3} + x + C \), where \( C \) is the constant of integration. So, the general solution is: \[ y = \frac{x^3}{3} + x + C \]

Apply initial condition to find the particular solution

Use the initial condition \( y = 1 \) at \( x = 1 \) to find \( C \). Substitute \( x = 1 \) and \( y = 1 \) into the general solution:\[ 1 = \frac{1^3}{3} + 1 + C \]Simplify:\[ 1 = \frac{1}{3} + 1 + C \]\[ 1 = \frac{4}{3} + C \]Subtract \( \frac{4}{3} \) from both sides:\[ C = 1 - \frac{4}{3} = -\frac{1}{3} \]

Write the particular solution

Substitute \( C = -\frac{1}{3} \) back into the general solution:\[ y = \frac{x^3}{3} + x - \frac{1}{3} \]This is the particular solution that satisfies the given condition.

Key concepts

Initial Condition

An initial condition in a differential equation provides specific values for the variable and its derivatives at a particular point. This condition helps to determine the specific solution to a differential equation. In our exercise, the initial condition is given by the statement that the function satisfies the relationship \( y = 1 \) when \( x = 1 \).
This information is crucial because differential equations typically have many possible solutions, referred to as families of solutions. The initial condition allows us to pinpoint which particular member of this family fits the specific problem we are attempting to solve.
By plugging \( x = 1 \) and \( y = 1 \) into the general solution, we can solve for the constant of integration \( C \). This process converts the general solution into a particular solution that meets the conditions of the problem.

Integration

Integration is a fundamental concept in solving differential equations. It essentially helps to "reverse" the process of differentiation. For this exercise, solving the differential equation involves integrating both sides with respect to \( x \).
The differential equation we are working with is \( \frac{dy}{dx} = x^2 + 1 \). To solve it, we integrate the right-hand side, \( \int (x^2 + 1) \, dx \).
During integration, each term is integrated separately, which means:

• \( x^2 \) integrates to \( \frac{x^3}{3} \)
• 1 integrates to \( x \)

This gives us \( y = \frac{x^3}{3} + x + C \), where \( C \) is the constant of integration arising from the indefinite integration process, indicating that there are potentially many solutions (differing by a constant).

Particular Solution

A particular solution to a differential equation satisfies the initial condition given in the problem. In our task, the solution must not only solve the differential equation but also fit the condition \( y = 1 \) at \( x = 1 \).
This is achieved by first finding a general solution: \( y = \frac{x^3}{3} + x + C \). To find the particular solution, we substitute \( x = 1 \) and \( y = 1 \) into this general equation:

• 1 = \( \frac{1^3}{3} + 1 + C \)

After solving for \( C \), we find that \( C = -\frac{1}{3} \).
The particular solution is then \( y = \frac{x^3}{3} + x - \frac{1}{3} \). This specific solution accounts for the given condition of the problem, making it unique to this scenario.

General Solution

A general solution represents the family of functions that solve a differential equation before accounting for any specific initial conditions. It includes an arbitrary constant \( C \) which allows it to capture an infinite set of possible solutions.
In this exercise, after integrating the equation \( \frac{dy}{dx} = x^2 + 1 \), we obtain the general solution:
\[ y = \frac{x^3}{3} + x + C \]
This equation describes all possible functions that fit the original differential equation. The constant \( C \) represents the numerous vertical shifts of the curve in a graph.
To narrow it down to the specific curve that fits a real-world problem or a specific scenario given by initial conditions, we find \( C \) by substituting the conditions into this general form. With the integration process creating a wave of possible outcomes, the initial condition makes the bridge to a specified, clear outcome.