/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 15 (a) Sketch some typical integral... [FREE SOLUTION] | 91Ó°ÊÓ

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(a) Sketch some typical integral curves of the differential equation \(y^{\prime}=y / 2 x\). (b) Find an equation for the integral curve that passes through the point \((2,1)\).

Short Answer

Expert verified
The equation for the integral curve through (2,1) is \( y = \frac{1}{\sqrt{2}} x^{1/2} \).

Step by step solution

01

Understand the Differential Equation

The given differential equation is \(y' = \frac{y}{2x}\). This is a first-order linear separable differential equation that represents the slope of the tangent to the integral curves at any point \((x, y)\) in the plane.
02

Separate Variables

To solve this equation, we need to separate the variables \(x\) and \(y\). We can do this by rewriting the equation: \[ dy = \frac{y}{2x} dx \]Then rearrange the terms to isolate \(y\) and its differential on one side and \(x\) and its differential on the other:\[ \frac{dy}{y} = \frac{1}{2x} dx \]
03

Integrate Both Sides

Now, integrate both sides of the equation:\[ \int \frac{dy}{y} = \int \frac{1}{2x} dx \]This results in:\[ \ln |y| = \frac{1}{2} \ln |x| + C \]where \(C\) is the constant of integration.
04

Solve for y

Next, solve for \(y\) using exponentiation to eliminate the logarithms:\[ |y| = e^{\frac{1}{2} \ln |x| + C} \]which simplifies to:\[ |y| = e^{C} \cdot (|x|^{1/2}) \]Let \(C_1 = e^C\), then version both sides eliminates the absolute value function assuming a principal value (positive):\[ y = C_1 \cdot x^{1/2} \].
05

Find Specific Integral Curve Through Given Point

We find \(C_1\) using the point \((2, 1)\): Substitute \(x = 2\) and \(y = 1\) into \(y = C_1 \cdot x^{1/2}\):\[ 1 = C_1 \cdot 2^{1/2} \]Solving for \(C_1\), we have \(C_1 = \frac{1}{\sqrt{2}}\).
06

Write the Specific Integral Curve Equation

Substitute \(C_1\) back into the equation for \(y\):\[ y = \frac{1}{\sqrt{2}} \cdot x^{1/2} \] which represents the specific integral curve through the point \((2, 1)\).

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Key Concepts

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

Integral Curves
Integral curves are a fundamental concept when dealing with differential equations. They represent the family of curves that satisfy a given differential equation. Each curve illustrates how the function behaves across different conditions in the plane. For instance, if we consider the differential equation given by \(y' = \frac{y}{2x}\), we're essentially looking at how the slope, \(y'\), changes based on not just \(x\) but also \(y\).
When sketching integral curves, we interpret the given differential equation to understand how each curve might look. These curves are generated by the solution of the differential equation, and each specific curve corresponds to a set of initial conditions. In the exercise, sketching involves finding integral curves that meet specified conditions. Specifically, these curves give us visual insights into how solutions to the differential equation morph as initial conditions vary. Integral curves provide an intuitive graphic representation of solutions to differential equations and often help verify analytical solutions.
Separable Differential Equation
A separable differential equation is a type of differential equation in which variables can be separated and rewritten such that all terms involving one variable are on one side of the equation and all terms involving another variable are on the opposite side. The ability to separate variables is crucial because it allows us to integrate each side independently.
For the differential equation \(y' = \frac{y}{2x}\), separation of variables involves rewriting it as \(\frac{dy}{dx} = \frac{y}{2x}\). We can rearrange this to \(\frac{dy}{y} = \frac{1}{2x} dx\). By separating the equation in this way, each side is now ready for integration, leading to individual integrals that can be solved step-by-step.
This technique is particularly useful because it simplifies complex problems into more manageable ones. Once separated, each integral can be calculated, eventually leading to a general solution for the given differential equation. Separable differential equations are commonly encountered in many fields of science and engineering, making this method a valuable tool for solving a variety of practical problems.
Constant of Integration
When integrating to solve a differential equation, a constant of integration often appears as part of the solution. This constant arises because indefinite integration yields a family of functions, each differing by a constant. For instance, when integrating \(\int \frac{dy}{y} = \ln |y|\), we add an arbitrary constant \(C\) to account for these infinite possibilities.
In the steps given for solving the differential equation, after integration, the solution was \(\ln |y| = \frac{1}{2} \ln |x| + C\). This \(C\) is the constant of integration that ensures all potential solutions are accounted for. The constant is crucial because it allows us to pinpoint specific solutions based on initial conditions—like finding which particular curve passes through a specified point \((2, 1)\).
Determining the value of the constant of integration involves substituting known values into the solved equation to find a particular solution. This value, once calculated, refines the general solution into one unique to the initial conditions provided. In the exercise, solving \(1 = C_1 \cdot 2^{1/2}\) led to finding \(C_1 = \frac{1}{\sqrt{2}}\), thus enabling the precise integral curve equation, \(y = \frac{1}{\sqrt{2}} \cdot x^{1/2}\). Such specificity is necessary for accurately describing the behavior of a system modeled by the differential equation.

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Most popular questions from this chapter

The water in a polluted lake initially contains 1 lb of mercury salts per \(100,000\) gal of water. The lake is circular with diameter \(30 \mathrm{m}\) and uniform depth \(3 \mathrm{m}\). Polluted water is pumped from the lake at a rate of \(1000 \mathrm{gal} / \mathrm{h}\) and is replaced with fresh water at the same rate. Construct a table that shows the amount of mercury in the lake (in \(1 \mathrm{b})\) at the end of each hour over a 12 -hour period. Discuss any assumptions you made. [Note: Use \(1 \mathrm{m}^{3}=264\) gal.]

A slope field of the form \(y^{\prime}=f(y)\) is said to be \(autonomous.\) (a) Explain why the tangent segments along any horizontal line will be parallel for an autonomous slope field. (b) The word autonomous means "independent." In what sense is an autonomous slope field independent? (c) Suppose that \(G(y)\) is an antiderivative of \(1 /[f(y)]\) and that \(C\) is a constant. Explain why any differentiable function defined implicitly by \(G(y)-x=C\) will be a solution to the equation \(y^{\prime}=f(y).\)

Solve the differential equation by the method of integrating factors. $$ \frac{d y}{d x}+2 x y=x $$

In each part, verify that the functions are solutions of the differential equation by substituting the functions into the equation. $$ \begin{array}{l}{y^{\prime \prime}-4 y^{\prime}+4 y=0} \\ {\text { (a) } e^{2 x} \text { and } x e^{2 x}} \\ {\text { (b) } c_{1} e^{2 x}+c_{2} x e^{2 x}\left(c_{1}, c_{2} \text { constants }\right)}\end{array} $$

(a) Prove that solutions need not be unique for nonlinear initial-value problems by finding two solutions to $$ y \frac{d y}{d x}=x, \quad y(0)=0 $$ (b) Prove that solutions need not exist for nonlinear initial-value problems by showing that there is no solution for $$ y \frac{d y}{d x}=-x, \quad y(0)=0 $$

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