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Chapter 8: Problem 46

In Problems 46-54, solve each differential equation with the given initial condition. $$ \frac{d y}{d x}=2 \frac{y}{x}, \text { with } y(1)=1 $$

Short Answer

Expert verified
The solution is \( y = x^2 \).

Step by step solution

01

- Rearrange the differential equation

The given differential equation is \( \frac{d y}{d x} = 2 \frac{y}{x} \). Notice that this is a separable differential equation. To solve it, we first rewrite it as \( \frac{d y}{y} = 2 \frac{d x}{x} \), which separates the variables.
02

- Integrate both sides

Integrate both sides of the equation: \( \int \frac{1}{y} \, d y = \int 2 \frac{1}{x} \, d x \). The left side integrates to \( \ln |y| \) and the right side integrates to \( 2 \ln |x| + C \), where \( C \) is the integration constant. Thus, we have \( \ln |y| = 2 \ln |x| + C \).
03

- Exponentiate to solve for \( y \)

Exponentiate both sides of \( \ln |y| = 2 \ln |x| + C \) to solve for \( y \). This gives \( |y| = e^C \cdot x^2 \). We can write \( y = A x^2 \) where \( A = \pm e^C \), representing a constant.
04

- Use the initial condition

Use the initial condition \( y(1) = 1 \) to find \( A \). Substituting \( x = 1 \) and \( y = 1 \) into \( y = A x^2 \) gives \( 1 = A \cdot 1^2 \), so \( A = 1 \). Thus, \( y = x^2 \).

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Key Concepts

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

Separable Differential Equations
Separable differential equations are a special type of differential equations where you can separate the variables. This makes them relatively simple to solve compared to other types. The key idea is to manipulate the equation such that all terms involving one variable (let's say it's \( y \)) are on one side of the equation, and all terms involving another variable (such as \( x \)) are on the other side. This allows each side to be integrated separately.

For example, consider the differential equation from the exercise: \( \frac{d y}{d x} = 2 \frac{y}{x} \). You can re-arrange it to be \( \frac{d y}{y} = 2 \frac{d x}{x} \). Now, terms involving \( y \) and \( x \) are separated, which is the essence of a separable differential equation.

To solve these equations:
  • Rearrange the equation to separate the variables.
  • Integrate both sides separately with respect to their respective variables.
  • Solve for the variable in question, often using additional information like initial conditions.
This method leverages integration as a stepping stone to the solution, making it a straightforward process.
Initial Conditions
Initial conditions are crucial in differential equations because they allow us to find particular solutions. Without them, we would only have a general solution which includes an arbitrary constant. This constant can represent an infinite number of solutions. The initial condition helps to 'pin down' which one of these infinite solutions is relevant to the specific scenario we are looking at.

In this exercise, the initial condition given is \( y(1) = 1 \). This tells us that when \( x = 1 \), \( y \) must be equal to 1. By substituting \( x = 1 \) and \( y = 1 \) into the equation \( y = A x^2 \), we can solve for \( A \). It turns out \( A = 1 \), so our particular solution is \( y = x^2 \).

Initial conditions are essential for:
  • Specifying a unique solution to a differential equation.
  • Making the abstract solution concrete and applicable to real-world situations.
  • Testing the solution manually by plugging in values to ensure accuracy.
They transform general solutions into specific answers tailored to the problem at hand.
Integration
Integration is a fundamental technique in calculus used for finding a function based on its derivative. In the context of differential equations, integration is employed to solve equations where derivatives are given and you need to find the original function.

For separable differential equations, once the equation is rearranged to separate the variables, both sides can be integrated with respect to their respective variables. Using our example, after separating the variables: \( \int \frac{1}{y} \, dy = \int 2 \frac{1}{x} \, dx \), the integration gives us the general solution: \( \ln |y| = 2 \ln |x| + C \). Here, \( C \) represents the integration constant.

Integration is used:
  • To reverse the process of differentiation, giving us the original function from its derivative.
  • To uncover the solutions of differential equations, critical in fields like physics and engineering.
  • To incorporate initial conditions, which allow the conversion from general to particular solutions.

    Through integration, the seemingly complex process of solving differential equations becomes manageable, enabling us to uncover valuable insights into natural and mathematical phenomena.

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Most popular questions from this chapter

In our compartment model we assumed that inflows and outflows are matched at \(q\) to keep the volume of water in the tank constant. It's often useful when modeling, for example, the flow of pollutant into a pristine environment, to consider what can occur if the inflows and outflows do not match. Let's assume that the tank initially contains a volume \(V_{0}\) of water. Water flows into the tank at rate \(q_{\mathrm{in}}\), and out of the tank at rate \(q_{\text {out. }}\) (You may assume \(q_{\text {in }}>q_{\text {out } .}\) ) Suppose that the water flowing into the tank contains a concentration \(C_{I}\) of solute. As usual we write \(C(t)\) for the concentration in the tank. (a) Show that the concentration in the tank can be modeled using a differential equation: $$ \frac{d}{d t}(C V)=q_{\text {in }} C_{I}-q_{\text {out }} C $$ (b) Previously we were able to treat \(V\) as a constant. Now \(V\) changes with time. Derive a formula for \(V(t)\). (c) By substituting your formula for \(V(t)\) into (a), derive a differential equation for \(C(t)\). (d) In general we cannot analyze the behavior of the solution \(C(t)\) using techniques from Section \(8.2 .\) Why not? (e) Let's assume \(C_{l}=0\). Then show that your equation from (c) can be written as: $$ \frac{d C}{d t}=\frac{-q_{\mathrm{in}} C}{V_{0}+\left(q_{\mathrm{in}}-q_{\mathrm{out}}\right) t} $$ (f) Assume some definite values for the constants in \((8.57):\) \(q_{\mathrm{in}}=2, q_{\text {out }}=1\), and \(V_{0}=20 .\) Assuming \(C(0)=1\), solve \((8.57)\) to find \(C(t) .\) Show that \(\lim _{t \rightarrow \infty} C(t)=0\).

By breaking down each equation into two parts that you can sketch, determine how many equilib\mathrm{\\{} r i a ~ e a c h ~ d i f f e r e n t i a l ~ e q u a t i o n ~ h a s , ~ a n d ~ c l a s s i f y ~ t h e m ~ a s ~ s t a b l e ~ or unstable. You do not need to determine the location of the equilibria. $$ \frac{d N}{d t}=1-N-N^{3} $$

Subpopulation Interactions in Patchy Habitats To derive our model for patchy habitat we assumed that a fixed fraction, \(m\), of occupied sites became extinct in each unit of time. Often, however the survival of the population at a site depends on the number of subpopulations in the surrounding sites. If different subpopulations compete for limited resources, then the per site mortality rate may not be a constant, but may increase with \(p\) because, as \(p\) increases, competition between subpopulations increases. In questions 13 and 14 we will study the effect of different models for competition between subpopulations. The term \(p^{2}\) describes the density-dependent extinction of patches; that is, the per-patch extinction rate is \(p\), and a fraction \(p\) of patches are occupied, resulting in patches going extinct at a total rate of \(p^{2}\). The colonization of vacant patches is the same as in the Levins model. Then the fraction of occupied patches obeys a differential equation: $$ \frac{d p}{d t}=c p(1-p)-p^{2} $$ where \(c>0\). (a) Show that there are two possible equilibrium values for \(p\) in \([0,1]\) (which you should calculate) and determine their stability. (b) Does the patch model always predict a nontrivial equilibrium when \(c>0\) ? Contrast with what we found for the Levins model in Section 8.3.2.

For Problems 13-28 make vector field plots of each of the differential equations. Find any equilibria of each differential equation and use your vector field plot to classify whether each equilibrium is stable or unstable. $$ \frac{d y}{d t}=y-1 $$

Find the equilibria of the following differential equations. $$ \frac{d y}{d t}=y^{1 / 3}-1 $$
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