Problem 1 - Calculus Early Transcendentals

Question

Confirm that \(y=3 e^{x^{3}}\) is a solution of the initial-value problem \(y^{\prime}=3 x^{2} y, y(0)=3\)

Step-by-Step Solution

Answer

Yes, \(y = 3 e^{x^3}\) is a solution to the problem.

Step 1: Differentiate the given solution

To confirm if \(y = 3 e^{x^3}\) is a solution to the differential equation, first find its derivative with respect to \(x\). Utilize the chain rule for differentiation. The derivative is \(y' = \frac{d}{dx} (3 e^{x^3}) = 3 \cdot e^{x^3} \cdot \frac{d}{dx} (x^3)\). Since \(\frac{d}{dx} (x^3) = 3x^2\), the expression becomes \(y' = 3 \cdot e^{x^3} \cdot 3x^2 = 9x^2 e^{x^3}\).

Step 2: Verify the differential equation

Substitute \(y = 3 e^{x^3}\) and \(y' = 9x^2 e^{x^3}\) into the differential equation \(y' = 3x^2 y\). The right-side of the equation is \(3x^2 \cdot y = 3x^2 \cdot (3 e^{x^3}) = 9x^2 e^{x^3}\). This matches \(y'\), confirming that the function satisfies the differential equation.

Step 3: Verify the initial condition

Check if the function satisfies the initial condition \(y(0) = 3\). Substitute \(x = 0\) into the function: \(y(0) = 3 e^{0^3} = 3 e^{0} = 3 \cdot 1 = 3\). This confirms the function satisfies the initial condition at \(x = 0\).

Step 4: Conclusion

Since \(y = 3 e^{x^3}\) satisfies both the differential equation and the initial condition, it is confirmed as a solution to the initial-value problem.

Key Concepts

Initial Value Problem

An initial value problem (IVP) involves solving a differential equation with a given initial condition. This means you need to find a function that not only fulfills the differential equation but also meets an initial criterion at a specific point. In our problem, the differential equation is \(y' = 3x^2 y\), and the initial condition is \(y(0) = 3\). This initial condition ensures that the function respects certain boundary requirements, making it unique and fully determined. Solving an IVP typically involves two steps: verifying that the solution to the differential equation is correct and checking that it matches the initial condition.

Derivatives

Derivatives are a fundamental concept in calculus, representing the rate of change of a function with respect to a variable. In our exercise, we start by differentiating the function \(y = 3e^{x^3}\).
This involves calculating the derivative \(y'\), which describes how \(y\) changes as \(x\) changes. The goal was to use this derivative to check if the initial function is a solution to the given differential equation. Differentiating functions correctly is crucial; any errors here will lead to incorrect solutions or ineffective verification. In this particular example, differentiating required using the chain rule, a technique often essential for handling composite functions.

Chain Rule

The chain rule is a powerful tool in calculus used for differentiating composite functions. Whenever you have a function inside another function, the chain rule helps you take derivatives.
Here's how it was applied in our problem: for \(y = 3e^{x^3}\), we needed to differentiate \(e^{x^3}\). The chain rule helps by first differentiating the outer function \(e^u\) (where \(u = x^3\)) and then multiplying by the derivative of the inner function \(u = x^3\).
This results in \(3e^{x^3} imes 3x^2 = 9x^2 e^{x^3}\). Understanding the chain rule can simplify complex problems since it breaks down the differentiation process into easier steps, especially when dealing with exponential functions.

Exponential Functions

Exponential functions involve exponents that are themselves variables. This makes their rate of growth quite remarkable and allows them to model many natural phenomena like population growth or radioactive decay. In our problem, we handle the function \(y = 3e^{x^3}\), where \(e\) is the base of the natural logarithm.
Exponential functions are unique because their rate of change is proportional to the function’s current value, a property that played a key part in verifying the solution to the differential equation. Because \(e^{x^3}\) changes rapidly as \(x\) increases, it's important to manipulate such functions carefully, respecting their properties, as seen in the differentiation step. Exponential functions are omnipresent in calculus and differential equations.