Problem 109 - University Calculus: Early Transcendentals

Question

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

Step-by-Step Solution

Answer

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

Step 1: Integrate the Third Derivative

The given differential equation is \( \frac{d^{3} y}{d x^{3}} = 6 \). Integrate this with respect to \( x \) to get the second derivative \( y''(x) \). Integrating gives \( y''(x) = 6x + C_1 \), where \( C_1 \) is a constant of integration.

Step 2: Apply the Second Derivative Initial Condition

We know from the initial conditions that \( y''(0) = -8 \). Using this, substitute \( x = 0 \) into the equation from Step 1: \( -8 = 6(0) + C_1 \). Solving for \( C_1 \), we find \( C_1 = -8 \). Therefore, \( y''(x) = 6x - 8 \).

Step 3: Integrate the Second Derivative

Now, integrate \( y''(x) = 6x - 8 \) with respect to \( x \) to find \( y'(x) \): \( y'(x) = \int (6x - 8) \, dx = 3x^2 - 8x + C_2 \), where \( C_2 \) is another constant of integration.

Step 4: Apply the First Derivative Initial Condition

Using the initial condition \( y'(0) = 0 \), substitute \( x = 0 \) into the equation from Step 3: \( 0 = 3(0)^2 - 8(0) + C_2 \). Solving for \( C_2 \), we find \( C_2 = 0 \). Therefore, \( y'(x) = 3x^2 - 8x \).

Step 5: Integrate the First Derivative

Integrate \( y'(x) = 3x^2 - 8x \) with respect to \( x \) to find \( y(x) \): \( y(x) = \int (3x^2 - 8x) \, dx = x^3 - 4x^2 + C_3 \), where \( C_3 \) is another constant of integration.

Step 6: Apply the Function Initial Condition

Using the initial condition \( y(0) = 5 \), substitute \( x = 0 \) into the equation from Step 5: \( 5 = (0)^3 - 4(0)^2 + C_3 \). Solving for \( C_3 \), we find \( C_3 = 5 \). Therefore, \( y(x) = x^3 - 4x^2 + 5 \).

Key Concepts

Third-Order Differential Equation

A third-order differential equation involves the third derivative of a function. This means we are looking for a function, say \(y(x)\), whose third derivative, \(\frac{d^3y}{dx^3}\), satisfies the given equation. In the context of the exercise, we had the equation \(\frac{d^3y}{dx^3} = 6\). This constant differential equation describes the constant rate of change of the rate of change of the rate of change of \(y\). To solve this, we need to integrate the equation multiple times. Each integration reduces the order of the derivative, bringing us closer to finding the original function, \(y(x)\). Third-order differential equations often require initial conditions at different derivative levels to find a particular solution.

Integration

Integration is the reverse process of differentiation. It's a fundamental method to solve differential equations, especially when dealing with derivatives. In the given problem, the first step was to integrate the third derivative of \(y(x)\), \(\frac{d^3y}{dx^3} = 6\), to find the second derivative, \(y''(x)\). Every time we integrate, an arbitrary constant (constant of integration) is introduced because the derivative of a constant is zero. So, it can’t be determined from differentiation alone. For this reason, when we integrated to find \(y''(x)\), we wrote:

• \(y''(x) = 6x + C_1\)

The steps are repeated through successive integrations to find both the first derivative and finally the function itself.

Constants of Integration

When integrating a function, such as from \(\frac{d^3y}{dx^3}\) to \(y''(x)\), a constant of integration is added to the result. These constants are crucial for identifying specific solutions in initial value problems. In our exercise:

• \(C_1\) was added after the first integration, resulting in \(y''(x) = 6x + C_1\).


• \(C_2\) was added to find the first derivative after integrating \(y''(x)\), giving \(y'(x) = 3x^2 - 8x + C_2\).


• Finally, \(C_3\) was added after integrating \(y'(x)\) to produce the function \(y(x) = x^3 - 4x^2 + C_3\).

Initial conditions help in solving these constants, tailoring the solution to the specific problem.

Initial Conditions

Initial conditions are vital in defining a unique solution to a differential equation that otherwise has infinitely many solutions due to the constants of integration. They are values given for functions and their derivatives at specific points, used to solve for these constants. For instance, in our problem:

• The condition \(y''(0) = -8\) was used to solve for \(C_1\).


• Similarly, \(y'(0) = 0\) allowed us to find \(C_2\).


• Finally, \(y(0) = 5\) helped in determining \(C_3\).

These conditions ensure that the solution fits the initial starting behavior described in the problem, leading to a well-defined and precise solution.