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Chapter 11: Problem 21

(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 \(n ?\)

Short Answer

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

Step by step solution

01

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: \[\frac{dy}{dx} = \frac{d}{dx}(C x^{n}) = Cn x^{n-1}.\]
02

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:\[C n x^{n} - 3 C x^{n} = 0.\]
03

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:\[n - 3 = 0.\]
04

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 \( n = 3 \).
05

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 \( C \):\[C = \frac{40}{8} = 5.\]
06

Conclusion

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

In Problems \(55-58,\) give an example of: An expression for \(f(x)\) such that the differential equation \(d y / d x=f(x)+x y-\cos x\) is separable.

Give an example of: A differential equation for any quantity which grows in two ways simultaneously: on its own at a rate proportional to the cube root of the amount present and from an external contribution at a constant rate.

(a) A cup of coffee is made with boiling water and stands in a room where the temperature is \(20^{\circ} \mathrm{C}\) If \(H(t)\) is the temperature of the coffee at time \(t,\) in minutes, explain what the differential equation $$\frac{d H}{d t}=-k(H-20)$$ says in everyday terms. What is the sign of \(k ?\) (b) Solve this differential equation. If the coffee cools to \(90^{\circ} \mathrm{C}\) in 2 minutes, how long will it take to cool to \(60^{\circ} \mathrm{C}\) degrees?

The amount of radioactive carbon- 14 in a sample is measured using a Geiger counter, which records each disintegration of an atom. Living tissue disintegrates at a rate of about 13.5 atoms per minute per gram of carbon. In 1977 a charcoal fragment found at Stonehenge, England, recorded 8.2 disintegrations per minute per gram of carbon. Assuming that the half-life of carbon- 14 is 5730 years and that the charcoal was formed during the building of the site, estimate the date at which Stonehenge was built.

Consider a conflict between two armies of \(x\) and \(y\) soldiers, respectively. During World War I, F. W. Lanchester assumed that if both armies are fighting a conventional battle within sight of one another, the rate at which soldiers in one army are put out of action (killed or wounded) is proportional to the amount of fire the other army can concentrate on them, which is in turn proportional to the number of soldiers in the opposing army. Thus Lanchester assumed that if there are no reinforcements and \(t\) represents time since the start of the battle, then \(x\) and \(y\) obey the differential equations $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. In this problem we adapt Lanchester's model for a conventional battle to the case in which one or both of the armies is a guerrilla force. We assume that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (a) Give a justification for the assumption that the rate at which a guerrilla force is put out of action is proportional to the product of the strengths of the two armies. (b) Write the differential equations which describe a conflict between a guerrilla army of strength \(x\) and a conventional army of strength \(y,\) assuming all the constants of proportionality are 1 (c) Find a differential equation involving \(d y / d x\) and solve it to find equations of phase trajectories. (d) Describe which side wins in terms of the constant of integration. What happens if the constant is zero? (e) Use your solution to part (d) to divide the phase plane into regions according to which side wins.
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