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Chapter 8: Problem 56

In Problems you will need to solve differential equations by separation of variables. In these problems it will not always be possible to solve explicitly for \(y\) in terms of \(t ;\) instead your solution may take the form of an implicit function relating the two variables. $$ \frac{d y}{d t}=\frac{y t}{\ln y} \text { where } y(1)=e $$

Short Answer

Expert verified
The implicit solution to the differential equation is \( (\ln y)^2 = t^2 \).

Step by step solution

01

Separate the Variables

First, we need to rewrite the given differential equation, \( \frac{d y}{d t}=\frac{y t}{\ln y} \), to separate the variables \( y \) and \( t \). This can be done by dividing both sides by \( y \) and multiplying both sides by \( \ln y \), yielding: \( \ln y \frac{d y}{y} = t \, dt \).
02

Integrate Both Sides

Next, integrate both sides of the equation from Step 1. The left-hand side becomes \( \int \ln y \frac{d y}{y} \), and the right-hand side becomes \( \int t \, dt \).
03

Solve the Left-Hand Side Integration

For the left-hand side, use substitution. Let \( u = \ln y \), then \( du = \frac{1}{y} \, dy \). The integral becomes \( \int u \, du \), which results in \( \frac{u^2}{2} = \frac{(\ln y)^2}{2} \).
04

Solve the Right-Hand Side Integration

For the right-hand side, integrate with respect to \( t \), thus \( \int t \, dt = \frac{t^2}{2} + C \), where \( C \) is the constant of integration.
05

Write the Combined Solution

Combine the results of Steps 3 and 4: \( \frac{(\ln y)^2}{2} = \frac{t^2}{2} + C \).
06

Apply Initial Condition

Use the initial condition \( y(1) = e \) to find \( C \). Substitute \( t = 1 \) and \( y = e \) into the equation: \( \frac{(\ln e)^2}{2} = \frac{1^2}{2} + C \). Since \( \ln e = 1 \), the equation simplifies to \( \frac{1}{2} = \frac{1}{2} + C \), giving \( C = 0 \).
07

Final Implicit Solution

Substitute \( C = 0 \) back into the combined solution to obtain the implicit solution: \( (\ln y)^2 = t^2 \).

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Key Concepts

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

Separation of Variables
Separation of variables is a powerful technique used to solve differential equations. The aim is to manipulate the equation so that each variable and its differential appear on different sides of the equation. This makes it easier to integrate each side separately. To perform separation of variables in the given differential equation \( \frac{d y}{d t} = \frac{y t}{\ln y} \), we rearrange terms to isolate \( y \) on one side and \( t \) on the other. Here are the steps:
  • Divide both sides by \( y \) to get \( \frac{d y}{y} = \frac{t}{\ln y} \, dt \).
  • Multiply both sides by \( \ln y \) to get \( \ln y \frac{d y}{y} = t \, dt \).
Now, each side of the equation is ready to be integrated separately. This careful separation sets the stage for the integration step, leading towards finding a solution to the differential equation.
Implicit Function
An implicit function is different from an explicit function in that it doesn't solve explicitly for one variable in terms of the others. Instead, it relates variables within an equation form. For the solved differential equation problem, our solution came out as an implicit function: \( (\ln y)^2 = t^2 \). In this form, neither \( y \) nor \( t \) is expressed solely in terms of the other variable. Solving implicitly is common in differential equations when simplifying the solution to an explicit form isn鈥檛 straightforward or necessary. It allows us to express the relationship between the variables correctly without further simplification that could be complex or impossible. This form expresses all possible solutions where the initial condition is also respected.
Initial Condition
An initial condition is a specified value that a solution to a differential equation must satisfy. Using an initial condition is crucial as it helps to find the specific solution from a family of possible solutions. In the problem, the initial condition given is \( y(1) = e \). This means when \( t = 1 \), \( y \) must equal \( e \). To use this condition:
  • Substitute \( t = 1 \) and \( y = e \) into the implicit solution \( (\ln y)^2 = t^2 \).
  • This gives \( (\ln e)^2 = 1^2 \), simplifying to \( 1 = 1 + C \). Here, we solve for \( C \) which turns out to be \( 0 \).
Applying the initial condition narrows down the infinite set of solutions to the one specific solution that meets the problem's criteria. This makes it unique and ensures it adheres to the initial state described by the problem.

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Most popular questions from this chapter

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable. $$ \frac{d N}{d t}=\frac{N-1}{N+1} \quad N \geq 0 $$

Find all equilibria, and, by calculating the eigenvalue of the differential equation, determine which equilibria are stable and which are unstable.$$ \frac{d y}{d x}=\frac{y^{2}-y}{y^{2}+1} $$

For Problems \(49-56\) determine whether the equilibrium at \(x=0\) is stable, unstable, or semi-stable. $$ \frac{d x}{d t}=x^{3} $$

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.
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