Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

First find the general solution (involving a constant C) for the given differential equation. Then find the particular solution that satisfies the indicated condition. (See Example 2.) \(\frac{d y}{d x}=x^{2}+1 ; y=1\) at \(x=1\)

Step-by-Step Solution

Answer

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

Step 1: Identify the Differential Equation

The given differential equation is \(\frac{dy}{dx} = x^2 + 1\). Our task is to find the general solution first.

Step 2: Integrate Both Sides

To solve for the general solution, integrate both sides of the equation with respect to \(x\). On the left side, integrate \(dy\), and on the right side, integrate \(x^2 + 1\). Thus, we have \(\int dy = \int (x^2 + 1) dx\).

Step 3: Perform the Integration

The integration yields \(y = \frac{x^3}{3} + x + C\), where \(C\) is the constant of integration. This represents the general solution to the differential equation.

Step 4: Apply the Initial Condition

We are given that \(y = 1\) when \(x = 1\). Substitute these values into the general solution to find \(C\). Substitute \(1 = \frac{1^3}{3} + 1 + C\).

Step 5: Solve for the Constant C

Simplify the equation \(1 = \frac{1}{3} + 1 + C\). This simplifies to \(1 = \frac{4}{3} + C\). Therefore, \(C = 1 - \frac{4}{3} = -\frac{1}{3}\).

Step 6: Write the Particular Solution

Substitute \(C = -\frac{1}{3}\) back into the general solution. The particular solution is \(y = \frac{x^3}{3} + x - \frac{1}{3}\).

Key Concepts

General Solution

In differential equations, finding the general solution is often your first goal. This solution encompasses all possible solutions by including an arbitrary constant, generally represented by \(C\). In this problem, our differential equation is \(\frac{dy}{dx} = x^2 + 1\).

To find the general solution, we integrate both sides of the equation. Integers or expressions on the right side, when integrated, will yield functions of \(x\) plus the constant \(C\). For our equation, the integration produces \(y = \frac{x^3}{3} + x + C\).

This equation includes \(C\) to account for solutions passing through different points in the plane. Essentially, the general solution represents a family of curves, all differing by vertical shifts dictated by \(C\).

Particular Solution

A particular solution takes the general solution and applies an initial condition to find a specific value for the constant \(C\). This gives a solution valid for a specific situation or set of data.

In this exercise, we apply the initial condition \(y = 1\) at \(x = 1\) to our general solution \(y = \frac{x^3}{3} + x + C\). By substituting the values \(y = 1\) and \(x = 1\) into the equation, we solve for \(C\):

• 1 = \(\frac{1}{3} + 1 + C\)
• Solving gives \(C = -\frac{1}{3}\)

Substituting \(C\) back, we obtain the particular solution \(y = \frac{x^3}{3} + x - \frac{1}{3}\), which meets the provided initial condition perfectly.

Integration

Integration is a core process used in solving differential equations to find functions that satisfy them. Essentially, integrating means finding a function whose derivative is the given function.

In this exercise, we integrate \(x^2 + 1\) with respect to \(x\). The integral of \(x^2\) is \(\frac{x^3}{3}\) and the integral of \(1\) is \(x\). Thus, the integration results in \(\frac{x^3}{3} + x\), albeit with a constant \(C\) included to account for the indefinite integral.

The process can be summarized as:

• Set up the equation: \(\int (x^2 + 1) \, dx = y\)
• Solve: \(y = \frac{x^3}{3} + x + C\)

Integration transforms the derivative back to its original function plus a constant, making it indispensable in finding general solutions.

Initial Condition

Initial conditions are vital for determining particular solutions to differential equations. They provide specific values for the variables, which help fix the constant \(C\) in the general solution.

In our exercise, the condition \(y = 1\) when \(x = 1\) acts as the necessary initial condition. By substituting these values into the general solution \(y = \frac{x^3}{3} + x + C\), we are able to isolate and solve for \(C\).

This process ensures the solution fits the real-world or discussed scenario. Here's how we apply an initial condition:

• Start with: \(1 = \frac{1}{3} + 1 + C\)
• Solve to find: \(C = -\frac{1}{3}\)

Applying the initial condition here helps decode the arbitrary constant, tailoring the general solution into one that specifically fits this scenario.