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Chapter 11: Problem 15

use separation of variables to find the solution to the differential equation subject to the initial condition. $$2 \frac{d u}{d t}=u^{2}, u(0)=1$$

Short Answer

Expert verified
The solution is \( u(t) = \frac{-1}{\frac{1}{2}t - 1} \).

Step by step solution

01

Rewrite the Differential Equation

The given differential equation is \(2 \frac{du}{dt} = u^2\). To use separation of variables, first rewrite it as \(\frac{du}{dt} = \frac{u^2}{2}\). This equation is now ready for separation of variables.
02

Separate Variables

Rearrange the equation to separate the variables \(u\) and \(t\). We have:\[ \frac{du}{u^2} = \frac{1}{2} dt. \]This separates the differential equation into parts only involving \(u\) on one side and \(t\) on the other.
03

Integrate Both Sides

Integrate both sides of the equation:\[ \int \frac{du}{u^2} = \int \frac{1}{2} dt. \]The left side integrates to \(-\frac{1}{u}\), and the right side integrates to \(\frac{1}{2}t + C\), where \(C\) is a constant of integration.
04

Solve for the Constant Using Initial Condition

Apply the initial condition \(u(0) = 1\) to find \(C\):\[ -\frac{1}{1} = \frac{1}{2}(0) + C \implies C = -1. \]
05

Solve for \( u(t) \)

Substitute \(C = -1\) back into the integrated equation:\[ -\frac{1}{u} = \frac{1}{2}t - 1. \]Solving for \(u(t)\), we get:\[ u(t) = \frac{-1}{\frac{1}{2}t - 1}. \]

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Key Concepts

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

Differential Equations
Differential equations involve relationships between a function and its derivatives. They describe how a quantity changes with respect to another variable. In this problem, our differential equation is \(2 \frac{du}{dt} = u^2\). Here,
  • \(u\) is a function of \(t\), the independent variable.
  • \(\frac{du}{dt}\) represents the rate of change of \(u\) with respect to \(t\).
Differential equations can model real-world phenomena such as population growth, chemical reactions, or mechanical vibrations. Solving them often gives insight into how these systems evolve over time, helping us predict future behavior or understand past events.
Integration
Integration is a core mathematical tool used to find functions given their derivatives. In this exercise, after separating the variables, we integrate both sides of the equation:
  • Left side: \( \int \frac{du}{u^2} \) integrates to \(-\frac{1}{u}\).
  • Right side: \( \int \frac{1}{2} dt \) results in \(\frac{1}{2}t + C\), where \(C\) is a constant.
Through integration, we uncover the relationship between \(u\) and \(t\). This step translates the problem from seeing how one variable changes, into one where we have an actual expression linking them. By integrating, we're essentially reversing the differentiation process.
Initial Condition
An initial condition gives a specific value for the solution of a differential equation at a particular point. In our problem, we have the condition \(u(0) = 1\). This means when \(t = 0\), \(u\) must equal 1.
  • Use this condition: Substitute \(t = 0\) and \(u(0) = 1\) into the equation after integrating, to find the constant \(C\).
  • Importance: Using the initial condition ensures the solution fits the specific scenario described by the problem.
Without accounting for the initial condition, our solution would remain broad, representing a family of possible solutions. Applying \(u(0) = 1\) pinpoints the exact solution that meets this initial behavior.
Separable Equations
Separable equations are a type of differential equation where the variables can be separated onto different sides of the equation. For this exercise, we deal with
  • Left side: Terms involving \(u\).
  • Right side: Terms involving \(t\).
The original equation \(2 \frac{du}{dt} = u^2\) is rewritten as \(\frac{du}{dt} = \frac{u^2}{2}\), then further arranged into \(\frac{du}{u^2} = \frac{1}{2} dt\). This clear separation allows each side to be integrated independently.
  • Why it matters: Separable equations often appear in physics and engineering because they simplify the integration process.
  • Solving approach: After separating, we can integrate each side separately and find relationships between the variables.
Separation of variables transforms complex differential equations into simpler forms, making them easier to handle and solve.

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Most popular questions from this chapter

Give an example of: A graph of \(d P / d t\) against \(P\) if \(P\) is a logistic function which increases when \(020\).

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 \(\lim _{x \rightarrow \infty} g(x)=\infty,\) then \(\lim _{x \rightarrow \infty} f(x)=\infty.\)

An item is initially sold at a price of \(\$ p\) per unit. Over time, market forces push the price toward the equilibrium price, \(\$ p^{*},\) at which supply balances demand. The Evans Price Adjustment model says that the rate of change in the market price, \(\$ p,\) is proportional to the difference between the market price and the equilibrium price. (a) Write a differential equation for \(p\) as a function of \(t.\) (b) Solve for \(p.\) (c) Sketch solutions for various different initial prices, both above and below the equilibrium price. (d) What happens to \(p\) as \(t \rightarrow \infty ?\)

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 \(g(0)=1\) and \(x(x)\) is increasing for \(x \geq 0,\) then \(f(x)\) is also increasing for \(x \geq 0.\)

As you know, when a course ends, students start to forget the material they have learned. One model called the Ebbinghaus model) assumes that the rate at which a student forgets material is proportional to the difference between the material currently remembered and some positive constant, \(a\) (a) Let \(y=f(t)\) be the fraction of the original material remembered \(t\) weeks after the course has ended. Set up a differential equation for \(y .\) Your equation will contain two constants; the constant \(a\) is less than \(y\) for all \(t.\) (b) Solve the differential equation. (c) Describe the practical meaning (in terms of the amount remembered) of the constants in the solution \(y=f(t).\)
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