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Families of curves often arise as solutions of differential equations. Match the families of curves with the differential equations of which they are solutions. (a) \(\frac{d y}{d x}=\frac{y}{x}\) (I) \(y=x e^{k x}\) (b) \(\frac{d y}{d x}=k y\) (II) \( y=x^{k}\) (c) \(\frac{d y}{d x}=k y+\frac{y}{x}\) (III) \(y=e^{2 x}\) (d) \(\frac{d y}{d x}=\frac{k y}{x}\) (IV) \( y=k x\)

Short Answer

Expert verified
(a) IV, (b) III, (c) I, (d) II

Step by step solution

01

Identify Type of Differential Equation

We have four differential equations that need to be matched to their corresponding family of curves. These are given as: (a) \( \frac{d y}{d x}=\frac{y}{x} \), (b) \( \frac{d y}{d x}=k y \), (c) \( \frac{d y}{d x}=k y+\frac{y}{x} \), and (d) \( \frac{d y}{d x}=\frac{k y}{x} \).
02

Analyze First Differential Equation

Let's analyze equation (a) \( \frac{d y}{d x}=\frac{y}{x} \), which is separable. If separated and integrated, it yields \( y = Cx \). This suggests the family of curves is of the form \( y = kx \), matching (IV).
03

Analyze Second Differential Equation

Equation (b) is \( \frac{d y}{d x}=k y \), which is a simple first-order linear separable differential equation. The solution is \( y = Ce^{kx} \), matching family (III) as it resembles \( y = e^{2x} \) if we assume \( C = 1 \) and \( k = 2 \).
04

Analyze Third Differential Equation

Equation (c) is \( \frac{d y}{d x}=k y+\frac{y}{x} \), can also be solved using an integrating factor or another method as it is a non-homogeneous differential equation. Its solution can be derived and one of the solutions is \( y = xe^{kx} \), which matches (I).
05

Analyze Fourth Differential Equation

The final equation, (d) \( \frac{d y}{d x}=\frac{k y}{x} \), when separated and integrated, reveals the solution \( y = Cx^k \), matching the family of curves given by (II).
06

Compile Matches

Thus, the correct matches are as follows: (a) with (IV) "\( y = kx \)" (b) with (III) "\( y = e^{2x} \)" (c) with (I) "\(y = xe^{kx}\)" (d) with (II) "\( y = x^k \)".

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Key Concepts

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

Separable Differential Equations
A separable differential equation is a specific type of differential equation where the two variables can be separated on different sides of the equation. This means you can express the equation in the form \( \frac{dy}{dx} = g(x)h(y) \). By "separable," we mean that you can rearrange it so that one side only involves \( y \) and its derivatives and the other side only involves \( x \). This separation is beneficial for solving the differential equation by integration.

To solve a separable equation like \( \frac{dy}{dx} = \frac{y}{x} \), you would rearrange to get \( \frac{dy}{y} = \frac{dx}{x} \). Then, integrate both sides:
  • \( \int \frac{1}{y} \, dy = \int \frac{1}{x} \, dx \)
  • The results are \( \ln|y| = \ln|x| + C \),
where \( C \) is the constant of integration. Exponentiating both sides gives \( y = Cx \), aligning with what the equation suggests as the family of solutions (like "(d)": \( y = kx \), where \( k \) can be any constant).

This method is efficient for solving many first-order differential equations where variables can be isolated and integrated separately.
Integrating Factor
An integrating factor is a technique mainly used to solve linear first-order differential equations that cannot be easily separated. These equations are usually in the form \( \frac{dy}{dx} + P(x)y = Q(x) \). The integrating factor, which is often denoted as \( \mu(x) \), is obtained by computing \( e^{\int P(x) \, dx} \).

This factor simplifies the differential equation, making it solvable by converting it into an exact equation. For instance, take equation (c) from the problem: \( \frac{dy}{dx} = ky + \frac{y}{x} \). Rearranging, it becomes \( \frac{dy}{dx} - ky = \frac{y}{x} \). The coefficient of \( y \) would suggest an integrating factor of \( e^{\int -k \, dx} \), which is \( e^{-kx} \).

Multiplying the whole equation by this integrating factor can transform it into one that can be integrated easily:
  • The equation simplifies to a form \( \frac{d}{dx}[\mu(x)y] = \mu(x)Q(x) \).

By integrating both sides, you can solve for \( y \). Thus, with the correct integrating factor, complex differential equations become easier to manage and solve.
First-Order Linear Differential Equations
First-order linear differential equations form a class of equations that are linear with respect to the variable \( y \) and its first derivative. These equations generally take a standard form:
  • \( \frac{dy}{dx} + P(x)y = Q(x) \).

The beauty of first-order linear differential equations lies in their systematic method of solution. This includes the use of integrating factors, as discussed previously, or in some cases, separation of variables if applicable.

For example, consider the differential equation \( \frac{dy}{dx} = ky \), known as the exponential growth equation. It can be solved by easily separating variables and integrating on both sides:
  • the integration process leads to \( y = Ce^{kx} \),
reflecting phenomena like population growth or radioactivity decay.

Moreover, first-order differential equations are foundational, as they set the stage for solving more complex higher-order equations and systems used in numerous fields such as physics, engineering, and economic modeling.

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Most popular questions from this chapter

Is the statement true or false? Assume that \(y=f(x)\) is a solution to the equation \(d y / d x=g(x) .\) If the statement is true, explain how you know. If the statement is false, give a counterexample. If \(\lim _{x \rightarrow \infty} g(x)=0,\) then \(\lim _{x \rightarrow \infty} f(x)=0.\)

Give an example of: A differential equation that has a logarithmic function as a solution.

Solve the initial value problem. $$y^{\prime \prime}+6 y^{\prime}+10 y=0, \quad y(0)=0, \quad y^{\prime}(0)=0$$

If \(y=e^{2 t}\) is a solution to the differential equation $$ \frac{d^{2} y}{d t^{2}}-5 \frac{d y}{d t}+k y=0 $$ find the value of the constant \(k\) and the general solution to this equation.

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 $$ \begin{array}{l} \frac{d x}{d t}=-a y \\ \frac{d y}{d t}=-b x \quad a, b>0 \end{array} $$ Near the end of World War II a fierce battle took place between US and Japanese troops over the island of Iwo Jima, off the coast of Japan. Applying Lanchester's analysis to this battle, with \(x\) representing the number of US troops and \(y\) the number of Japanese troops, it has been estimated \(^{35}\) that \(a=0.05\) and \(b=0.01\) (a) Using these values for \(a\) and \(b\) and ignoring reinforcements, write a differential equation involving \(d y / d x\) and sketch its slope field. (b) Assuming that the initial strength of the US forces was 54,000 and that of the Japanese was 21,500 draw the trajectory which describes the battle. What outcome is predicted? (That is, which side do the differential equations predict will win?) (c) Would knowing that the US in fact had 19,000 reinforcements, while the Japanese had none, alter the outcome predicted?

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