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Q10 E

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Fundamentals Of Differential Equations And Boundary Value Problems

Found in: Page 14

Book edition 9th

Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider

Pages 616 pages

ISBN 9780321977069

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Short Answer

In Problems 9-13, determine whether the given relation is an implicit solution to the given differential equation. Assume that the relationship implicitly defines y as a function of x and use implicit differentiation.
$y - \log_{e} y = x^{2} + 1$ , $\frac{dy}{dx} = \frac{2 xy}{y - 1}$

The given relation is an implicit solution to the given differential equation.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step1: Differentiating the given relation

As, in the given relation $y - \log_{e} y = x^{2} + 1$ , y is defined implicitly as the function of x, so by using implicit differentiation, we will differentiate the given relation concerning x,

$\frac{d}{dx} (y - \log_{e} y) = \frac{d}{dx} (x^{2} + 1) \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = 2 x$

Step 2: Simplification of the differential equation obtained in step 1

$(1 - \frac{1}{y}) \frac{dy}{dx} = 2 x (\frac{y - 1}{y}) \frac{dy}{dx} = 2 x \frac{dy}{dx} = \frac{2 xy}{y - 1}$

Which is identical to the given differential equation.

Thus, the relation $y - \log_{e} y = x^{2} + 1$ is an implicit solution to the differential equation $\frac{dy}{dx} = \frac{2 xy}{y - 1}$ .

Most popular questions for Math Textbooks

Mixing. Suppose a brine containing 0.2 kg of salt per liter runs into a tank initially filled with 500 L of water containing 5 kg of salt. The brine enters the tank at a rate of 5 L/min. The mixture, kept uniform by stirring, is flowing out at the rate of 5 L/min (see Figure 2.6).

(a)Find the concentration, in kilograms per liter, of salt in the tank after 10 min. [Hint: Let A denote the number of kilograms of salt in the tank at t minutes after the process begins and use the fact that

rate of increase in A =rate of input - rate of exit.

A further discussion of mixing problems is given in Section 3.2.]

(b) After 10 min, a leak develops in the tank and an additional liter per minute of mixture flows out of the tank (see Figure 2.7). What will be the concentration, in kilograms per liter, of salt in the tank 20 min after the leak develops? [Hint: Use the method discussed in Problems 31 and 32.]

In Problems 14–24, you will need a computer and a programmed version of the vectorized classical fourth-order Runge–Kutta algorithm. (At the instructor’s discretion, other algorithms may be used.)†

Using the vectorized Runge–Kutta algorithm with h = 0.5, approximate the solution to the initial value $3 t^{2} y'' - 5 ty' + 5 y = 0; y (1) = 0, y' (1) = \frac{2}{3}$ problemat t = 8.

Compare this approximation to the actual solution $y (t) = t^{\frac{5}{3}} - t$ .

Let $ϕ (x)$ denote the solution to the initial value problem

$\frac{dy}{dx} = x - y, y (0) = 1$

⦁ Show that $ϕ (x) = 1 - ϕ' (x) = 1 - x + ϕ (x)$

⦁ Argue that the graph of $ϕ$ is decreasing for x near zero and that as x increases from zero, $ϕ (x)$ decreases until it crosses the line y = x, where its derivative is zero.

⦁ Let x* be the abscissa of the point where the solution curve $y = ϕ (x)$ crosses the line $y = x$ .Consider the sign of $ϕ (x *)$ and argue that $ϕ$ has a relative minimum at x*.

⦁ What can you say about the graph of $y = ϕ (x)$ for x > x*?

⦁ Verify that y = x – 1 is a solution to $\frac{dy}{dx} = x - y$ and explain why the graph of $y = ϕ (x)$ always stays above the line $y = x - 1$ .

⦁ Sketch the direction field for $\frac{dy}{dx} = x - y$ by using the method of isoclines or a computer software package.

⦁ Sketch the solution $y = ϕ (x)$ using the direction field in part (f).

In problems 1-4 Use Euler’s method to approximate the solution to the given initial value problem at the points $x = 0.1, 0.2, 0.3, 0.4$ , and $0.5$ , using steps of size $0.1 (h = 0.1)$ .

$\frac{d y}{d x} = \frac{x}{y}, y (0) = - 1$

The motion of a set of particles moving along the x‑axis is governed by the differential equation $\frac{dx}{dt} = t^{3} - x^{3},$ where $x (t)$ denotes the position at time t of the particle.

⦁ If a particle is located at $x = 1$ when $t = 1$ , what is its velocity at this time?

⦁ Show that the acceleration of a particle is given by $\frac{d^{2} x}{{dt}^{2}} = 3 t^{2} - 3 t^{3} x^{2} + 3 x^{5} .$

⦁ If a particle is located at $x = 2$ when $t = 2.5$ , can it reach the location $x = 1$ at any later time?

[Hint: $t^{3} - x^{3} = (t - x) (t^{2} + xt + x^{2}) .$ ]

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