Problem 23 - Calculus and its applications

Question

(a) find the particular solution of each differential equation as determined by the initial condition, and (b) check the solution by substituting into the differential equation. \(y^{\prime}=x^{2}+2 x-3 ; \quad y=4\) when \(x=0\)

Step-by-Step Solution

Answer

The particular solution is \(y = \frac{x^3}{3} + x^2 - 3x + 4\), matching the given differential equation and initial condition.

Step 1: Identify the Differential Equation and Initial Condition

The given differential equation is \(y' = x^2 + 2x - 3\). The initial condition is \(y = 4\) when \(x = 0\). This means that when we solve the differential equation, we need to find a particular solution that satisfies \(y(0) = 4\).

Step 2: Integrate the Differential Equation

To find the particular solution, integrate the right-hand side of the equation with respect to \(x\): \[ y = \int (x^2 + 2x - 3) \, dx = \frac{x^3}{3} + x^2 - 3x + C \]where \(C\) is the constant of integration.

Step 3: Apply the Initial Condition

Use the initial condition \(y(0) = 4\) to find \(C\). Substitute \(x = 0\) and \(y = 4\) into the integrated function:\[ 4 = \frac{(0)^3}{3} + (0)^2 - 3(0) + C \]Simplifying, we find \(C = 4\). So the particular solution is:\[ y = \frac{x^3}{3} + x^2 - 3x + 4 \]

Step 4: Verify the Solution

Differentiate the particular solution \(y = \frac{x^3}{3} + x^2 - 3x + 4\) with respect to \(x\) to find \(y'\):\[ y' = x^2 + 2x - 3 \]This is the original differential equation, confirming that our solution is correct.

Key Concepts

Understanding Initial Conditions

In differential equations, an initial condition is a value that helps us find a specific solution out of many general solutions. When you integrate a differential equation, it usually includes an arbitrary constant, known as the constant of integration. Without the initial condition, you end up with a family of solutions. The initial condition acts like a guiding star, steering us to a particular path in the vast sky of possible solutions.
In the exercise provided, you have the differential equation \(y' = x^2 + 2x - 3\) with the initial condition \(y = 4\) when \(x = 0\). This crucial piece of information tells us the exact point on the solution curve:

• At \(x = 0\), the function \(y\) should equal 4.

So essentially, the initial condition is what allows us to pinpoint the exact curve from an infinite set of possibilities.

Explaining the Particular Solution

The particular solution is the unique solution derived from a differential equation, using the initial condition. This is achieved through integrating the equation and then applying the given initial condition to find the constant, which we often call \(C\).
In our exercise, we started by integrating the equation, resulting in a general solution: \[ y = \frac{x^3}{3} + x^2 - 3x + C. \] To find the particular solution, we used the initial condition where \(y = 4\) when \(x = 0\). Substituting these values helped us determine that \(C = 4\).

• Therefore, the particular solution is given as \( y = \frac{x^3}{3} + x^2 - 3x + 4 \).

This process narrows down the infinite solutions to the one that fits your specific problem. The particular solution is thus personalized for the condition provided.

Deciphering the Constant of Integration

The constant of integration, denoted \(C\), plays a crucial role in finding solutions to indefinite integrals and differential equations. It's essentially a placeholder for the unknown value that gets fixed once you apply the initial condition. Each time you solve a differential equation, the integration step yields this constant, signifying that many solutions could fit before being narrowed down.
When solving the differential equation \(y' = x^2 + 2x - 3\), the integration step leads to \[ y = \frac{x^3}{3} + x^2 - 3x + C. \] Here, \(C\) is indeterminate until you apply the specific initial condition. After substituting \(x = 0\) and \(y = 4\), you get:

• The equation \(4 = 0 + C\) simplifies to give \(C = 4\).

Thus, the constant is determined, allowing you to specify the particular solution. Without determining \(C\), you'd be left with a general solution that could represent numerous potential curves.