Problem 21 - Calculus

Question

(a) For what values of \(C\) and \(n\) (if any) is \(y=C x^{n}\) a solution to the differential equation $$x \frac{d y}{d x}-3 y=0 ?$$ (b) If the solution satisfies \(y=40\) when \(x=2,\) what more (if anything) can you say about \(C\) and \(n ?\)

Step-by-Step Solution

Answer

(a) \(n = 3\); (b) \(C = 5\).

Step 1: Differentiate the function

Consider the function \( y = C x^{n} \). First, we need to find its derivative with respect to \( x \). The derivative is obtained as follows: \[\frac{dy}{dx} = \frac{d}{dx}(C x^{n}) = Cn x^{n-1}.\]

Step 2: Substitute into the differential equation

Next, substitute \( \frac{dy}{dx} = Cn x^{n-1} \) and \( y = Cx^{n} \) into the differential equation:\[x \frac{dy}{dx} - 3y = xc n x^{n-1} - 3 C x^{n} = 0.\] This simplifies to:\[C n x^{n} - 3 C x^{n} = 0.\]

Step 3: Simplify the equation

Factor out \( C x^{n} \) from the equation:\[C x^{n} (n - 3) = 0.\] Since \( C x^{n} eq 0 \) (as \( x eq 0 \)), the solution to this equation is simply:\[n - 3 = 0.\]

Step 4: Solve for n

From \( n - 3 = 0 \), we solve for \( n \) and find:\[n = 3.\] Thus, the solution \( y = C x^{3} \) satisfies the given differential equation for \( n = 3 \).

Step 5: Apply initial condition to find C

Given the condition \( y = 40 \) when \( x = 2 \), use the equation \( y = C x^{3} \) to determine \( C \):\[40 = C (2)^{3} \40 = 8C.\] Solve for \( C \):\[C = \frac{40}{8} = 5.\]

Step 6: Conclusion

With the initial condition, we determine \( n = 3 \) and \( C = 5 \). Therefore, the specific solution to the differential equation is \( y = 5 x^{3} \).

Key Concepts

Understanding Derivatives

In calculus, the concept of a derivative is fundamental. It helps us determine how a function changes at any given point. Think of a derivative as a tool that gives us the slope of the function's curve at any specific location. For a function like \( y = C x^n \), computing the derivative involves using power rules. The result is \( \frac{dy}{dx} = Cn x^{n-1} \). Here, \( Cn \) is the coefficient, and \( x^{n-1} \) is the new power of \( x \), showing us the rate of change.

Derivatives play a critical part in solving differential equations. They allow us to substitute a function's rate of change into these equations to find solutions.

Deciphering Initial Conditions

Initial conditions specify the value of a function at a certain point, which is critical in determining specific solutions to differential equations. In our exercise, we have an initial condition given as \( y = 40 \) when \( x = 2 \). This tells us exactly where our solution must pass through on the coordinate plane.

By substituting \( x = 2 \) and \( y = 40 \) into the found general solution \( y = Cx^{n} \), we can solve for constants like \( C \) in the equation. Here, the initial condition ensures our particular solution aligns with this specific point.

Steps in Solution Verification

Solution verification ensures that the solution we've derived truly satisfies the original equation. Once we've proposed a solution, we need to substitute it back into the differential equation to check its validity.

In our task, we verified \( y = 5x^{3} \) by plugging back into the original: \( x\frac{dy}{dx} - 3y = 0 \). If it simplifies correctly, it confirms our solution is correct. Hence, it's crucial for confirming accuracy and reliability in problem-solving.

Approaches to Mathematical Problem Solving

Mathematical problem solving involves a systematic approach anchored in logic and calculation. When tackling differential equations like this exercise presents, you generally follow these steps:

• Identify the function and differentiate it appropriately.
• Substitute derivations into the original equation.
• Simplify the equation to uncover constants or variables.
• Use initial conditions to find specific values for these constants.
• Double-check by substituting back into the equation for verification.

This structured method ensures clarity and helps to unravel solutions efficiently.