Problem 79 - Thomas Calculus

Question

Solve the initial value problems in Exercises \(67-86\). $$ \frac{d^{2} y}{d x^{2}}=2-6 x ; \quad y^{\prime}(0)=4, \quad y(0)=1 $$

Step-by-Step Solution

Answer

The solution to the initial value problem is \( y(x) = x^2 - x^3 + 4x + 1 \).

Step 1: Integrate to Find the First Derivative

We start with the second derivative of the function: \( \frac{d^2 y}{dx^2} = 2 - 6x \). To find the first derivative, we integrate this with respect to \(x\). The integral of \(2\) is \(2x\), and the integral of \(-6x\) is \(-3x^2\). Thus, \[ y'(x) = 2x - 3x^2 + C \]where \(C\) is the constant of integration.

Step 2: Determine Constant from Initial Condition for the First Derivative

We use the given initial condition \(y'(0) = 4\) to find \(C\). Substitute 0 for \(x\) in the equation for \(y'(x)\):\[ 4 = 2(0) - 3(0)^2 + C \]So, \(C = 4\). This gives us:\[ y'(x) = 2x - 3x^2 + 4 \]

Step 3: Integrate to Find the Original Function

Now, integrate \(y'(x)\) with respect to \(x\) to find \(y(x)\). Integrating \(2x\) gives \(x^2\), integrating \(-3x^2\) gives \(-x^3\), and integrating \(4\) gives \(4x\). Thus,\[ y(x) = x^2 - x^3 + 4x + D \]where \(D\) is another constant of integration.

Step 4: Determine Constant from Initial Condition for the Original Function

Use the initial condition \(y(0)=1\) to find \(D\). Substitute 0 for \(x\) in the equation for \(y(x)\):\[ 1 = 0^2 - 0^3 + 4(0) + D \]Thus, \(D = 1\). Therefore, the original function is:\[ y(x) = x^2 - x^3 + 4x + 1 \]

Step 5: Confirm Solution Satisfies Initial Conditions

Check that this solution satisfies the initial conditions. The derivative \(y'(x) = 2x - 3x^2 + 4\) should equal 4 when \(x=0\), which it does:\[ y'(0) = 4 \]The function \(y(x) = x^2 - x^3 + 4x + 1\) should equal 1 when \(x=0\), which it does:\[ y(0) = 1 \]. This confirms the solution is correct.

Key Concepts

Second Order Differential Equations

A second order differential equation involves the second derivative of a function. It is expressed as \( \frac{d^{2} y}{d x^{2}} \) in our exercise. These equations are essential because they describe systems with acceleration or deceleration, such as moving cars or falling objects.

The given exercise involves solving such an equation where the second derivative \( \frac{d^2 y}{dx^2} = 2 - 6x \). Solving it requires finding a function \( y \) that satisfies this equation. This means we need to integrate twice to recover the original function starting from the second derivative.

Second order differential equations are quite prevalent in physics and engineering, dealing with anything from electrical circuits to spring motion. Understanding how to solve them helps in modeling real-world problems effectively.

Integration

Integration is a fundamental process in calculus that reverses differentiation. It is used to find functions from their derivatives.

In our problem, we start with the second derivative \( \frac{d^2 y}{dx^2} \) and integrate to find the first derivative \( y'(x) \). Here, integrating \( 2 \) gives \( 2x \), and integrating \( -6x \) results in \( -3x^2 \). Therefore, \( y'(x) = 2x - 3x^2 + C \).

By integrating the first derivative again, we reach the original function \( y(x) \). The integral of \( y'(x) = 2x - 3x^2 + 4 \) results in \( x^2 - x^3 + 4x + D \).

Integration allows us to step back from derivatives to the original function, essential in solving differential equations.

Constant of Integration

When integrating, we'll encounter an unknown constant, known as the "constant of integration."

This arises because differentiation removes constant terms, and we can't determine these purely from the integration process without additional information. That's why integrating a function results in a general solution that includes the constant \( C \) or \( D \) in our equations.

For example, after integrating for the first derivative \( y'(x) \), we get \( +C \), and again for \( y(x) \), another \( +D \) appears.

Initial conditions, like \( y'(0) = 4 \) and \( y(0) = 1 \), help us calculate the exact values of these constants, giving us a specific solution instead of a general one.

Initial Conditions

Initial conditions are special values provided in a problem to help find specific solutions to a differential equation.

These values tell us what the function or its derivatives should equal at particular points, like \( y'(0) = 4 \) or \( y(0) = 1 \). Applying these initial conditions helps determine the constants of integration we found previously as ambiguities in the general solution.

In the exercise, substituting the initial condition \( y'(0) = 4 \) identified the constant \( C \) as 4. Similarly, \( y(0) = 1 \) allowed the calculation of \( D \) as 1.

Using initial conditions is crucial as it converts a general solution into the one that fits the specific scenario being analyzed.