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Solve the differential equation. $$\frac { d u } { d r } = \frac { 1 + \sqrt { r } } { 1 + \sqrt { u } }$$

Short Answer

Expert verified
The general solution is \( u + \frac{2}{3} u^{3/2} = r + \frac{2}{3} r^{3/2} + C \).

Step by step solution

01

Rewrite the Differential Equation

The given differential equation is \( \frac{du}{dr} = \frac{1 + \sqrt{r}}{1 + \sqrt{u}} \). This can be separated as \( (1 + \sqrt{u}) du = (1 + \sqrt{r}) dr \).
02

Integrate Both Sides

Integrate both sides of the equation: \( \int (1 + \sqrt{u}) \, du = \int (1 + \sqrt{r}) \, dr \).
03

Solve the Left Integral

The left integral is \( \int (1 + \sqrt{u}) \, du \), which can be solved as \( u + \frac{2}{3} u^{3/2} + C_1 \).
04

Solve the Right Integral

The right integral is \( \int (1 + \sqrt{r}) \, dr \), which can be solved as \( r + \frac{2}{3} r^{3/2} + C_2 \).
05

Combine the Results

Combine the results from the integrals to form the equation \( u + \frac{2}{3} u^{3/2} = r + \frac{2}{3} r^{3/2} + C \), where \( C = C_2 - C_1 \).

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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 crucial technique for solving differential equations like the one we have here. It involves rearranging the equation so that all terms involving one variable are on one side of the equation, and all terms involving the other variable are on the other side. For the given problem, we have the differential equation \( \frac{du}{dr} = \frac{1 + \sqrt{r}}{1 + \sqrt{u}} \).
By applying separation of variables, we rewrite it as \((1 + \sqrt{u}) \, du = (1 + \sqrt{r}) \, dr \). This allows us to handle each variable separately, making the problem simpler to solve.
  • We treat \( du \) and \( dr \) as differentials and solve the equation by integrating each part.
  • The rearranged form makes it easier to manage complex expressions derived from the original equation.
This method is particularly useful for ordinary differential equations, where we work to isolate and integrate each side independently.
Integration
Integration is the process of finding the antiderivative of a function. In our example, after separating variables, we need to integrate both sides of the equation: \( \int (1 + \sqrt{u}) \, du = \int (1 + \sqrt{r}) \, dr \).
Integration allows us to "undo" the differentiation present in the equation, moving us closer to a solution. Each side of the separated equation is integrated separately:
  • The integral on the left is solved as \( u + \frac{2}{3} u^{3/2} + C_1 \).
  • The integral on the right is solved as \( r + \frac{2}{3} r^{3/2} + C_2 \).

Integration requires familiarity with basic antiderivatives and techniques like substitution. Once each integral is solved, we recombine our results. Integration is often an essential step in solving ordinary differential equations, especially those involving products or quotients of functions.
Integration Constants
When integrating, it's important to remember the constant of integration. Each indefinite integral introduces a constant because there are infinitely many antiderivatives for a given function. In our problem, the equations \( u + \frac{2}{3} u^{3/2} + C_1 \) and \( r + \frac{2}{3} r^{3/2} + C_2 \) both include their own constants, \( C_1 \) and \( C_2 \), respectively.
After integrating both sides, we combine the results to form the final equation:
  • \( u + \frac{2}{3} u^{3/2} = r + \frac{2}{3} r^{3/2} + C \)
  • The new constant \( C = C_2 - C_1 \)

This step is crucial because it represents the uncertainty in the initial conditions of the differential equation. Not accounting for integration constants can lead to incomplete solutions. They often help to adjust for the specific context or boundary conditions of the problem.
Ordinary Differential Equations
Ordinary differential equations (ODEs) involve functions of one independent variable and their derivatives. In our exercise, the equation \( \frac{du}{dr} = \frac{1 + \sqrt{r}}{1 + \sqrt{u}} \) is an ODE because it relates \( u \) and \( r \), and their derivatives.
These equations are foundational in mathematics, modeling various natural phenomena, from population dynamics to physical systems. ODEs can be solved by a variety of methods, including separation of variables, as we've used here.
  • Understanding how to manipulate and solve ODEs is fundamental for fields like engineering and physics.
  • They describe systems where quantities change relative to one another in a predictable manner.

Mastering ODEs allows students to apply mathematical concepts to real-world problems effectively, appreciating the details and nuances involved in dynamic systems.

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Most popular questions from this chapter

When a raindrop falls, it increases in size and so its mass at time \(t\) is a function of \(t,\) namely, \(m(t) .\) The rate of growth of the mass is \(k m(t)\) for some positive constant \(k .\) When we apply Newton's Law of Motion to the raindrop, we get \((m v)^{\prime}=g m,\) where \(v\) is the velocity of the raindrop (directed downward) and \(g\) is the acceleration due to gravity. The terminal velocity of the raindrop is \(\lim _{t \rightarrow \infty} v(t) .\) Find an expression for the terminal velocity in terms of \(g\) and \(k .\)

A system of differential equations is given. (a) Construct the phase plane, plotting all nullclines, labeling all equilibria, and indicating the direction of motion. (b) Obtain an expression for each equilibrium. \(z^{\prime}=z^{3}-4 z^{2}+3 z-2 w, \quad w^{\prime}=z-w-1\)

A system of differential equations is given. (a) Construct the phase plane, plotting all nullclines, labeling all equilibria, and indicating the direction of motion. (b) Obtain an expression for each equilibrium. \(p^{\prime}=p(1-p-q), \quad q^{\prime}=q(2-3 p-q), \quad p, q \geqslant 0\)

\(9-12\) Describe the motion of a particle with position \((x, y)\) as \(t\) varies in the given interval. \(x=3+2 \cos t, \quad y=1+2 \sin t, \quad \pi / 2 \leqslant t \leqslant 3 \pi / 2\)

Competition and cooperation Each system of differential equations is a model for two species that either compete for the same resources or cooperate for mutual benefit (flowering plants and insect pollinators, for instance). Decide whether each system describes competition or cooperation and explain why it is a reasonable model. (Ask yourself what effect an increase in one species has on the growth rate of the other.) $$\begin{aligned} \text { (a) } \frac{d x}{d t} &=0.12 x-0.0006 x^{2}+0.00001 x y \\ \frac{d y}{d t} &=0.08 x+0.00004 x y \\ \text { (b) } \frac{d x}{d t} &=0.15 x-0.0002 x^{2}-0.0006 x y \\ \frac{d y}{d t} &=0.2 y-0.00008 y^{2}-0.0002 x y \end{aligned}$$

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