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Chapter 11: Problem 24

Match solutions and differential equations. (Note: Each equation may have more than one solution.) (a) \(y^{\prime \prime}-y=0\) (b) \(x^{2} y^{\prime \prime}+2 x y^{\prime}-2 y=0\) (c) \(x^{2} y^{\prime \prime}-6 y=0\) (1) \(y=e^{x}\) (II) \(y=x^{3}\) (III) \(y=e^{-x}\) (IV) \(y=x^{-2}\)

Short Answer

Expert verified
(a) matches with (1) and (III); (b) matches with (IV); (c) matches with (II).

Step by step solution

01

Identify Solution for Equation (a)

The differential equation \(y'' - y = 0\) suggests a solution involving exponentials, as this is a homogeneous linear equation with constant coefficients. Testing the given solutions, we find that substituting \(y = e^{x}\) gives \(y'' = e^{x}\) and hence \(y'' - y = e^{x} - e^{x} = 0\), confirming \(y = e^{x}\) is a solution. Similarly, substituting \(y = e^{-x}\) gives \(y'' = e^{-x}\) and hence \(y'' - y = e^{-x} - e^{-x} = 0\), confirming \(y = e^{-x}\) is another solution.
02

Identify Solution for Equation (b)

The differential equation \(x^{2} y'' + 2x y' - 2y = 0\) suggests a solution involving powers of \(x\). Substituting \(y = x^{-2}\) yields \(y' = -2x^{-3}\) and \(y'' = 6x^{-4}\). Substituting into the equation gives: \(x^{2}(6x^{-4}) + 2x(-2x^{-3}) - 2x^{-2} = \)\[(6x^{-2} - 4x^{-2} - 2x^{-2}) = 0\]. This confirms that \(y = x^{-2}\) is a solution.
03

Identify Solution for Equation (c)

The differential equation \(x^{2} y'' - 6y = 0\) typically involves polynomial solutions. Testing \(y = x^{3}\), we find that \(y' = 3x^{2}\) and \(y'' = 6x\). Substituting into the equation, we get: \[x^{2}(6x) - 6(x^{3}) = 6x^{3} - 6x^{3} = 0\]. Thus, \(y = x^{3}\) satisfies the equation.

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Key Concepts

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

Solution Verification
Verifying a solution in the context of differential equations means checking whether a potential solution satisfies the given equation. It's a crucial step because it confirms that the solution fits the problem. Here鈥檚 how it works:
  • Take the proposed solution and compute any necessary derivatives.
  • Substitute these derivatives back into the original differential equation.
  • Simplify the equation to verify it results in an identity, often zero.
This process ensures that the solution is valid and consistent with the differential equation, enhancing understanding and preventing errors.
Homogeneous Equations
Homogeneous differential equations have all terms involving the unknown function or its derivatives with consistent degree, meaning no standalone constants or differing operations. In linear homogeneous equations like \(y'' - y = 0\), you鈥檒l often deal with solutions that are exponential functions. Recognizing homogeneity is key because it guides the types of assumed solutions:
  • Look for functions like \(e^{x}\) or \(x^{n}\) as they often fit.
  • Utilize characteristic equations for constant coefficient ones to find roots indicating solution forms.
These characteristics make solving homogeneous equations systematic, linking directly to the types of functions explored.
Exponential Functions
Exponential functions like \(e^{x}\) and \(e^{-x}\) play a pivotal role in solving linear homogeneous differential equations with constant coefficients. Their inherent property of retaining form under differentiation makes them ideal candidates:
  • The derivative of \(e^{x}\) is itself, simplifying substitution.
  • Exponential growth or decay in physical phenomena often modelled by such equations.
Knowing how to work with these functions enhances the solution verification process and allows for easily identifying valid solutions in many differential equations.
Polynomial Functions
Polynomial functions, represented as sums of powers of \(x\), like \(x^{3}\), are commonly examined when solving differential equations with variable coefficients. The characteristics that make them useful include:
  • Ease of differentiation鈥攅ach application of \(d/dx\) reduces the power by one, revealing patterns.
  • Appearing naturally in problems involving processes with proportional growth, like area or volume changes.
Solving equations like \(x^{2} y'' - 6y = 0\) benefits from this understanding, as testing polynomial solutions often leads to successful resolution.

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Most popular questions from this chapter

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.

Give an explanation for your answer. For any positive values of the constant \(k\) and any positive values of the initial value \(P(0),\) the solution to the differential equation \(d P / d t=k P(L-P)\) has limiting value \(L\) as \(t \rightarrow \infty\)

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 $$\begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array}$$. (a) For two armies of strengths \(x\) and \(y\) fighting a conventional battle governed by Lanchester's differential equations, write a differential equation involving \(d y / d x\) and the constants of attrition \(a\) and \(b\) (b) Solve the differential equation and hence show that the equation of the phase trajectory is $$a y^{2}-b x^{2}=C$$ for some constant \(C .\) This equation is called Lanchester's square law. The value of \(C\) depends on the initial sizes of the two armies.

Give the solution to the logistic differential equation with initial condition. $$\frac{d P}{d t}=0.04 P(1-0.0001 P) \text { with } P_{0}=200$$

Let \(L,\) a constant, be the number of people who would like to see a newly released movie, and let \(N(t)\) be the number of people who have seen it during the first \(t\) days since its release. The rate that people first go see the movie, \(d N / d t\) (in people/day), is proportional to the number of people who would like to see it but haven't yet. Write and solve a differential equation describing \(d N / d t\) where \(t\) is the number of days since the movie's release. Your solution will involve \(L\) and a constant of proportionality, \(k\).
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