Problem 24 - 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. \(y^{\prime}=3 x^{2}-x+5 ; \quad y=6\) when \(x=0\)

Step-by-Step Solution

Answer

The particular solution is \( y = x^3 - \frac{x^2}{2} + 5x + 6 \). Verification shows it satisfies the differential equation.

Step 1: Integrate the Differential Equation

To find the particular solution, we begin by integrating the differential equation. The equation given is \( y' = 3x^2 - x + 5 \). Integrated with respect to \( x \), it gives us:\[y = \int (3x^2 - x + 5) \, dx = \int 3x^2 \, dx - \int x \, dx + \int 5 \, dx.\]Calculating each integral individually:\[y = 3 \cdot \frac{x^3}{3} - \frac{x^2}{2} + 5x + C.\]Simplifying, we get:\[y = x^3 - \frac{x^2}{2} + 5x + C.\]

Step 2: Apply the Initial Condition

The initial condition provided is \( y = 6 \) when \( x = 0 \). Substitute these values into the integrated equation to find \( C \):\[6 = 0^3 - \frac{0^2}{2} + 5(0) + C \]\[6 = C.\]Thus, the particular solution is:\[y = x^3 - \frac{x^2}{2} + 5x + 6.\]

Step 3: Verify the Solution

To verify, substitute \( y = x^3 - \frac{x^2}{2} + 5x + 6 \) back into the original differential equation. First, find the derivative with respect to \( x \):\[y' = \frac{d}{dx} \left(x^3 - \frac{x^2}{2} + 5x + 6\right) = 3x^2 - x + 5.\]The derivative matches the original differential equation, confirming that the solution is correct.

Key Concepts

Particular Solution

A particular solution of a differential equation is a specific solution that satisfies the differential equation as well as initial or boundary conditions. When solving a differential equation, we often find a family of solutions that includes a constant of integration, known simply as 'C'. This family of solutions can satisfy the differential equation, but won't necessarily meet the given initial conditions.

• The role of the initial condition is to enable us to determine the specific value of 'C' that tailors the general solution to a particular solution.
• In our case, integrating the differential equation \( y' = 3x^2 - x + 5 \) gives us \( y = x^3 - \frac{x^2}{2} + 5x + C \).
• To pinpoint the particular solution, we use the initial condition \( y = 6 \) when \( x = 0 \). This allows us to find that \( C = 6 \), giving the particular solution \( y = x^3 - \frac{x^2}{2} + 5x + 6 \).

Initial Condition

The initial condition is a vital piece of the puzzle when solving differential equations, especially when determining a particular solution. It is a specific value given at a certain point and represents additional information external to the differential equation itself.

• In our example, the initial condition provided is \( y = 6 \) when \( x = 0 \).
• We substitute this value into the general integrated form of the equation to solve for the constant \( C \).
• This step effectively narrows down the infinitely many possible solutions to a single, unique particular solution.

The process is straightforward and mechanical but crucial, as it turns a general topic into a result that accurately represents the specific situation described by the initial condition.

Verification of Solution

Verification of a solution involves substituting the obtained solution back into the original differential equation to ensure it satisfies the equation. This step is essential to confirm our calculations.

• Once we have the particular solution, \( y = x^3 - \frac{x^2}{2} + 5x + 6 \), we compute its derivative.
• The derivative \( y' = 3x^2 - x + 5 \) is exactly the differential equation we started with, demonstrating correctness.
• This shows that all transformations and initial condition applications were executed correctly, resulting in a true solution.

Verification steps provide reassurance that the solution obtained is accurate and consistent with the original problem constraints.