Problem 87 - Thomas Calculus in SI Units

Question

Solve the initial value problems. $$\frac{d^{3} y}{d x^{3}}=6 ; \quad y^{\prime \prime}(0)=-8, \quad y^{\prime}(0)=0, \quad y(0)=5$$

Step-by-Step Solution

Answer

The solution is \( y(x) = x^3 - 4x^2 + 5 \).

Step 1: Integrate to Find the Second Derivative

Given the third derivative \( \frac{d^{3} y}{dx^{3}} = 6 \), our task is to integrate to find the second derivative. When integrating \( 6 \) with respect to \( x \), we get \( \frac{d^{2} y}{dx^{2}} = 6x + C_1 \), where \( C_1 \) is the constant of integration.

Step 2: Apply Initial Condition for the Second Derivative

We use the initial condition for the second derivative \( y''(0) = -8 \) to find \( C_1 \). Plugging in \( x = 0 \) gives \( 6(0) + C_1 = -8 \), so \( C_1 = -8 \). Thus, \( \frac{d^2 y}{dx^2} = 6x - 8 \).

Step 3: Integrate to Find the First Derivative

Now, integrate \( \frac{d^2 y}{dx^2} = 6x - 8 \) to find the first derivative. The integral of \( 6x \) is \( 3x^2 \), and the integral of \( -8 \) is \( -8x \), resulting in \( \frac{dy}{dx} = 3x^2 - 8x + C_2 \).

Step 4: Apply Initial Condition for the First Derivative

Using the initial condition \( y'(0) = 0 \), substitute \( x = 0 \) into the equation \( 3(0)^2 - 8(0) + C_2 = 0 \). This gives \( C_2 = 0 \). Thus, the first derivative is \( \frac{dy}{dx} = 3x^2 - 8x \).

Step 5: Integrate to Find the Function

Next, integrate \( \frac{dy}{dx} = 3x^2 - 8x \) to find \( y(x) \). The integral of \( 3x^2 \) is \( x^3 \), and the integral of \( -8x \) is \( -4x^2 \), so we have \( y(x) = x^3 - 4x^2 + C_3 \).

Step 6: Apply Initial Condition for the Function

Finally, use the initial condition \( y(0) = 5 \) to solve for \( C_3 \). Substitute \( x = 0 \) into the function: \( 0^3 - 4(0)^2 + C_3 = 5 \). Therefore, \( C_3 = 5 \). So the function is \( y(x) = x^3 - 4x^2 + 5 \).

Key Concepts

Differential Equations

A differential equation is an equation that involves an unknown function and its derivatives. In the context of our exercise, the differential equation is given by the third derivative, \( \frac{d^3 y}{dx^3} = 6 \). This means we are working with a third-order differential equation, indicating that we will need to integrate three times to find the original function \( y(x) \). Here, the order of the differential equation is crucial because it tells us the number of integration steps needed to reach the solution. Additionally, each integration introduces a constant of integration, which needs to be determined using given initial conditions. This process highlights the beauty and complexity of differential equations as they serve as a powerful tool to describe various phenomena across different fields.

Calculus

Calculus, the mathematics of change, plays a fundamental role in solving differential equations. It provides the framework for understanding how functions change, which is essential for finding the solutions to differential equations. The two main operations in calculus are differentiation and integration. Differentiation involves finding the derivative of a function, providing a mathematical way to describe how a function changes at any given point. Conversely, integration is the process of finding a function given its derivative. This duality is central to solving initial value problems, like the one in our exercise, as it helps reverse-engineer the unknown function from its derivative form back to the original form. Calculus is indispensable for solving complex systems in physics, engineering, and other sciences.

Integration

Integration is the process of finding a function whose derivative is known. In simpler terms, it is the reverse operation of differentiation. In our problem, starting with the third derivative \( \frac{d^3 y}{dx^3} = 6 \), integration is used to work backward to the original function \( y(x) \). Each integration step introduces a new constant, making it necessary to use initial conditions to find these constants. For example, integrating \( 6 \) gives us \( 6x + C_1 \). Repeating this process with the second and first derivatives eventually leads to the function itself. Integration is key to unraveling layers of derivatives to reconstruct the original function, thus solving the initial value problem.

Initial Conditions

Initial conditions are specific values of the function or its derivatives at a particular point, often given in initial value problems. These values are crucial as they allow us to determine the constants of integration introduced at each step. In our exercise, several initial conditions were given: \( y''(0) = -8 \), \( y'(0) = 0 \), and \( y(0) = 5 \).

• For the second derivative, \( y''(0) = -8 \) is used to find \( C_1 \).
• For the first derivative, \( y'(0) = 0 \) determines \( C_2 \).
• Finally, \( y(0) = 5 \) helps calculate \( C_3 \) for the function.

Applying these conditions ensures that the solution is not just mathematically consistent, but also specific to the problem's context. The use of initial conditions not only resolves the integration constants but also tailors the general solution to meet specific requirements set by the problem.