Problem 2 - Calculus Early Transcendentals

Question

Confirm that \(y=\frac{1}{4} x^{4}+2 \cos x+1\) is a solution of the initial- value problem \(y^{\prime}=x^{3}-2 \sin x, y(0)=3\)

Step-by-Step Solution

Answer

Yes, the function satisfies both the differential equation and the initial condition.

Step 1: Differentiate the Given Function

First, we need to find the derivative of the given function, \( y = \frac{1}{4}x^4 + 2\cos x + 1 \). To differentiate, we apply the power rule and the derivative of cosine:- The derivative of \( \frac{1}{4}x^4 \) is \( x^3 \).- The derivative of \( 2\cos x \) is \( -2\sin x \).- The derivative of the constant \( 1 \) is zero.Thus, \( y' = x^3 - 2\sin x \).

Step 2: Verify the Differential Equation

Now we need to confirm that the derivative \( y' = x^3 - 2\sin x \) satisfies the differential equation of the initial-value problem:\[ y' = x^3 - 2\sin x \]Since this matches the right-hand side of the given differential equation, \( y' = x^3 - 2\sin x \), the function satisfies the differential equation part.

Step 3: Check the Initial Condition

Finally, we verify the initial condition \( y(0) = 3 \) by substituting \( x = 0 \) into the original function:\[ y(0) = \frac{1}{4}(0)^4 + 2\cos(0) + 1 = 0 + 2(1) + 1 = 3 \]This shows that the initial condition is satisfied because \( y(0) = 3 \).

Key Concepts

Initial Value Problem

An Initial Value Problem (IVP) is a type of differential equation along with specific conditions, known as initial conditions. This tells you where to start solving the equation. In our exercise, the IVP is given as:

• Differential equation: \( y' = x^3 - 2\sin x \)
• Initial condition: \( y(0) = 3 \)

The goal is to find a function \( y(x) \) that satisfies both the differential equation and the initial condition. The initial condition ensures the uniqueness of the solution, meaning no two different paths can meet this same starting point. It determines a particular solution from a family of possible solutions to the differential equation. This is why, when solving differential equations, checking the initial condition is a crucial step.

Differentiation

Differentiation is the process of finding the derivative of a function. The derivative represents the rate of change of the function with respect to one of its variables. It's like measuring how fast something is changing at any given moment.
The main rules used in differentiation include the sum rule, product rule, chain rule, and power rule. In our given problem, to find the derivative of \( y = \frac{1}{4}x^4 + 2\cos x + 1 \), we apply the power rule for \( \frac{1}{4}x^4 \) and the rule for trigonometric functions for \( 2\cos x \). The constant 1 has a derivative of zero as constants don't change.
Effective differentiation is crucial in solving IVPs as it aids in verifying whether the proposed solution satisfies the differential equation.

Power Rule

The power rule is a basic but powerful tool in differentiation. It helps you quickly find the derivative of a term of the form \( ax^n \), where \( a \) is a constant, and \( n \) is a whole number.
The rule states:

• For \( f(x) = x^n \), the derivative \( f'(x) = nx^{n-1} \).

In the exercise, for the term \( \frac{1}{4}x^4 \), applying the power rule results in \( 4 \times \frac{1}{4}x^{4-1} = x^3 \).
This is an essential technique in solving and verifying differential equations because it enables you to find changes or slopes in polynomial functions efficiently.

Trigonometric Functions

Trigonometric functions like sine and cosine are essential in calculus. They describe periodic phenomena and work on angles, making them vital in various scientific and engineering fields.
Differentiation of trigonometric functions follows specific rules:

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \( -\sin x \).

In this exercise, the term \( 2\cos x \) differentiates to \( -2\sin x \). Understanding these derivatives is crucial when solving differential equations involving trigonometric functions.
Recognizing the behavior of these functions helps to ensure accurate calculations and results when incorporating them in both equations and solutions.