Problem 46 - Calculus Single Variable

Question

Solve the initial value problem \(y^{\prime}(x)=x g^{\prime}\left(x^{2}\right), g(4)=6\) \(y(2)=12\). Verify your formula with \(g(x)=x^{2}-2 x-2\).

Step-by-Step Solution

Answer

The solution is \( y(x) = \frac{1}{2}x^4 - x^2 + 8 \). After verifying, \( g \) confirms that \( y(2) = 12 \).

Step 1: Interpret the differential equation

We are given the initial value problem: \( y'(x) = x \cdot g'(x^2) \). This tells us that the derivative of \( y \) with respect to \( x \) is the product of \( x \) and the derivative of the function \( g \) evaluated at \( x^2 \). Our task is to find \( y(x) \).

Step 2: Understand given initial conditions

We are given two pieces of information: \( g(4) = 6 \) and \( y(2) = 12 \). The first condition will help in determining the function \( g \), and the second condition will be used to find the constant when integrating to find \( y(x) \).

Step 3: Analyze the function g

The task involves verifying our solution using \( g(x) = x^2 - 2x - 2 \). We need to compute \( g'(x) \), which is the derivative that will be used in the differential equation: \( g'(x) = 2x - 2 \).

Step 4: Substitute g' into the differential equation

Substitute \( g'(x^2) = 2x^2 - 2 \) into the given problem: \( y'(x) = x(2x^2 - 2) = 2x^3 - 2x \). This simplifies the differential equation to: \( y'(x) = 2x^3 - 2x \).

Step 5: Integrate to find y(x)

Integrate \( y'(x) = 2x^3 - 2x \) with respect to \( x \) to find \( y(x) \):\[y(x) = \int (2x^3 - 2x) \, dx = \frac{1}{2}(2x^4) - \frac{1}{2}(2x^2) + C = \frac{1}{2}x^4 - x^2 + C\]

Step 6: Use initial condition y(2)=12

Utilize the initial condition \( y(2) = 12 \) to solve for \( C \): \[12 = \frac{1}{2}(2)^4 - (2)^2 + C \]This becomes:\[12 = 8 - 4 + C\]\[C = 8 \]So, \( y(x) = \frac{1}{2}x^4 - x^2 + 8 \).

Step 7: Verify the formula

To verify, substitute back \( y(2)\) into the equation to ensure it equals 12:\[y(2) = \frac{1}{2}(2^4) - (2)^2 + 8 = 8 - 4 + 8 = 12 \]The verification confirms that the formula \( y(x) = \frac{1}{2}x^4 - x^2 + 8 \) satisfies the initial condition.

Key Concepts

Understanding Differential Equations

Differential equations are mathematical equations that involve derivatives, capturing the rate of change of a quantity. In our exercise, we are given a first-order differential equation:

• This equation expresses the relationship between the function \(y\) and its derivative \(y'(x)\).
• The equation we start with is \( y'(x) = x \cdot g'(x^2) \).
• This means that the rate of change of \( y \) with respect to \( x \) is dependent on both \( x \) and the derivative of another function \( g \) evaluated at \( x^2 \).

By analyzing these elements, we can better understand how to handle differential equations. They allow us to see how a function behaves based on its initial conditions and the way it changes over time or space.

In real-world applications, differential equations can describe phenomena such as motion, growth, decay, and much more, making them a fundamental concept in understanding mathematical modeling.

Mastering Integration Techniques

Integration is the reverse process of differentiation and is used to find a function given its derivative. In the context of our problem, integrating the expression \( y'(x) = 2x^3 - 2x \) helps us find \( y(x) \). The following steps outline how integration is performed:

• We recognize \( y'(x) = 2x^3 - 2x \) and set up the integral: \[y(x) = \int (2x^3 - 2x) \, dx = \int 2x^3 \, dx - \int 2x \, dx\]
• Calculating each part individually provides: \[y(x) = \frac{1}{2}(2x^4) - \frac{1}{2}(2x^2) + C = \frac{1}{2}x^4 - x^2 + C\]
• The result gives us the antiderivative of the function, also known as the primitive function. The constant \( C \) will be determined using the initial value \( y(2) = 12 \).

Remember, integration can sometimes involve specific techniques like substitution or integration by parts. For simpler polynomials like in our example, basic power rules suffice. Identifying the correct technique is crucial in solving differential equations effectively.

Verification of Solutions

Once a solution to a differential equation is obtained, verification ensures that the solution is correct and satisfies all given conditions. To verify the solution \( y(x) = \frac{1}{2}x^4 - x^2 + 8 \), we must:

• Substitute the solution back into the original differential equation to check consistency.
• Make sure that the initial conditions hold true when plugging them back into the solved equation.

In our example, we recheck the initial condition:

• By plugging \( x = 2 \) into \( y(x) \), \( y(2) = \frac{1}{2}(2^4) - (2)^2 + 8 = 8 - 4 + 8 \), which results in 12, confirming our calculations.

Verification serves as a safeguard, ensuring no mathematical steps were forgotten or incorrect. By consistently validating results, one gains confidence in the work and ensures that the solution accurately represents the modeled scenario.