Problem 105 - University Calculus: Early Transcendentals

Question

Solve the initial value problems in Exercises. $$\frac{d^{2} y}{d x^{2}}=2-6 x ; \quad y^{\prime}(0)=4, \quad y(0)=1$$

Step-by-Step Solution

Answer

The solution is \( y(x) = x^2 - x^3 + 4x + 1 \).

Step 1: Identify the Type of Differential Equation

The equation given is a second-order ordinary differential equation of the form \( \frac{d^{2} y}{dx^2} = 2 - 6x \). We need to integrate this equation twice to find the general solution for \( y(x) \).

Step 2: First Integration

Integrate the second derivative \( \frac{d^{2} y}{dx^2} = 2 - 6x \) with respect to \( x \) to find the first derivative \( y'(x) \). \[ \frac{d y}{dx} = \int (2 - 6x) \, dx = 2x - 3x^2 + C_1 \]Here, \( C_1 \) is the constant of integration.

Step 3: Apply Initial Condition to First Derivative

Use the initial condition \( y'(0) = 4 \) to find \( C_1 \). Substitute \( x = 0 \) and \( y'(0) = 4 \) into the equation derived in the previous step. \[ 4 = 2 \times 0 - 3 \times 0^2 + C_1 \rightarrow C_1 = 4 \]

Step 4: Second Integration

Integrate \( y'(x) = 2x - 3x^2 + 4 \) to find \( y(x) \). \[ y(x) = \int (2x - 3x^2 + 4) \, dx = x^2 - x^3 + 4x + C_2 \]Here, \( C_2 \) is another constant of integration.

Step 5: Apply Initial Condition to Find Final Solution

Use the initial condition \( y(0) = 1 \) to find \( C_2 \). Substitute \( x = 0 \) and \( y(0) = 1 \) into the equation derived above. \[ 1 = 0^2 - 0^3 + 4 \times 0 + C_2 \rightarrow C_2 = 1 \]Therefore, the solution \( y(x) \) is: \[ y(x) = x^2 - x^3 + 4x + 1 \]

Step 6: Verify the Solution with Initial Conditions

Verify that the solution satisfies the initial conditions. Substitute \( x = 0 \) into the solution:- For the first condition, \( y(0) = 1 \): \[ y(0) = 0 - 0 + 0 + 1 = 1 \] Confirmed as true.- For the second condition, \( y'(0) = 4 \): \[ y'(0) = 2 \times 0 - 3 \times 0^2 + 4 = 4 \] Confirmed as true.Both conditions are satisfied.

Key Concepts

Second-order differential equations

Second-order differential equations are critical in mathematics and physics as they model various real-world systems like oscillations and waves. These equations involve the second derivatives of a function, indicating how the rate of change itself changes. When solving, the equation is given in the form \( \frac{d^2 y}{dx^2} = f(x) \). To find a solution, we need to integrate this equation twice. This process gives us a general solution incorporating an unknown function \( y(x) \) and integration constants that define specific behavior or conditions a solution might follow.

One of the key features of second-order equations is their ability to express systems with memory or inertia, meaning the past state influences future behavior.
This aspect calls for careful use of initial conditions to ensure unique solutions are accurately deduced, ultimately satisfying a given problem, like the one presented.

Initial value problems

Initial value problems (IVPs) are a type of differential equation problem where the solution is determined based on specific initial conditions. These conditions usually include the value of the function and its derivatives at a particular point, such as \( y(0) = 1 \) and \( y'(0) = 4 \) in the solution process.

In solving IVPs, initial conditions play a crucial role in determining the particular solution from a family of all potential solutions after integrating the differential equation. They ensure that the solution aligns with physical realities or specific scenarios presented by the problem.

This specificity is essential not only in mathematics, but also in fields like engineering and physics, where predicting exact behaviors of systems under given states relies heavily on proper initial values.

Integration techniques

Integration techniques are mathematical methods used to find a function whose derivative is known. For solving differential equations like our given problem, integration is essential because it allows us to move from the second derivative of a function to the function itself, step by step.

In our exercise, after identifying the form of the differential equation, we start by integrating the right-hand side twice. The first integration step involves finding the first derivative \( y'(x) \), and we employ basic integral calculus:

• Integrate \( 2 - 6x \), yielding \( 2x - 3x^2 + C_1 \).

We apply the initial condition \( y'(0) = 4 \) to find \( C_1 \). Subsequent integration of \( y'(x) \) provides the solution's general form \( y(x) \).

The elegance of integration in such equations lies in its iterative nature: each step gradually uncovers more of the solution using fundamental integration skills, allowing us to ultimately find a precise expression for the system being studied.