Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. The equation \(u^{\prime}(x)=\left(x^{2} u^{7}\right)^{-1}\) is separable. b. The general solution of the separable equation \(y^{\prime}(t)=\frac{t}{y^{7}+10 y^{4}}\) can be expressed explicitly with \(y\) in terms of \(t\) c. The general solution of the equation \(y y^{\prime}(x)=x e^{-y}\) can be found using integration by parts.

Short Answer

Expert verified
In summary: a. The statement is true since the given equation is separable. b. The statement is false since an explicit solution for y(t) cannot be found. c. The statement is false since integration by parts is not necessary as the equation is separable.

Step by step solution

01

Statement a

To determine if the given equation \(u^{\prime}(x)=\left(x^{2} u^{7}\right)^{-1}\) is separable, we want to rewrite the equation as \(u'(x) = fu(x)\cdot g(x)\), where \(f\) is a function of \(x\) only, and \(g\) is a function of \(u\) only. Rewrite the equation as: \(u^{\prime}(x)=\frac{1}{x^{2} u^{7}}\) Indeed, the equation is separable as we have: \(u^{\prime}(x)=\frac{1}{x^{2}}\cdot\frac{1}{u^{7}}\) In this case, \(f(x) = \frac{1}{x^{2}}\) and \(g(u) = \frac{1}{u^{7}}\). Therefore, the statement is true.
02

Statement b

We are given a separable equation \(y^{\prime}(t)=\frac{t}{y^{7}+10y^{4}}\). In order to determine if an explicit solution can be found, we need to solve the equation by separation of variables. And see if we can solve for \(y(t)\) explicitly. Separate the variables: \(y'(t)(y^{7} + 10y^4) = t\) Integral the equation: \(\int y^{7} + 10y^4 dy = \int t dt\) Calculating the integrals, we have: \(\frac{1}{8}y^{8} + \frac{5}{5}y^5 + C = \frac{1}{2}t^2 + K\) It is evident that expressing the solution for \(y(t)\) explicitly is not possible. Thus, statement b is false.
03

Statement c

We are given the equation \(yy^{\prime}(x)=xe^{-y}\). In order to determine if integration by parts is an applicable method, we try to rewrite it in the form \(u\frac{dy}{dx}=v\), with the intention of choosing appropriate functions for integration by parts. However, if we do so, it appears that this equation is separable: \(\frac{dy}{dx}\frac{1}{y}= \frac{1}{x}e^{-y}\) Separate the variables: \(\frac{dy}{y} = e^{-y} \frac{dx}{x}\) We see that it is unnecessary to use integration by parts because the equation is separable. Assuming that integration by parts is needed for a non-separable equation only, statement c is false.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the initial value problem \(y^{\prime}(t)=-a y, y(0)=1,\) where \(a>0 ;\) it has the exact solution \(y(t)=e^{-a t},\) which is a decreasing function. a. Show that Euler's method applied to this problem with time step \(h\) can be written \(u_{0}=1, u_{k+1}=(1-a h) u_{k},\) for \(k=0,1,2, \ldots\) b. Show by substitution that \(u_{k}=(1-a h)^{k}\) is a solution of the equations in part (a), for \(k=0,1,2, \ldots .\) c. Explain why as \(k\) increases the Euler approximations \(u_{k}=(1-a h)^{k}\) decrease in magnitude only if \(|1-a h|<1\). d. Show that the inequality in part (c) implies that the time step must satisfy the condition \(0

Consider the first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=A,\) for \(t \geq 0,\) where \(a, b,\) and \(A\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to a horizontal line in the direction field. b. Draw a representative direction field in the case that \(a>0\) Show that if \(A>-b / a,\) then the solution increases for \(t \geq 0\) and if \(A<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\) Show that if \(A>-b / a,\) then the solution decreases for \(t \geq 0\) and if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

Solve the equation \(y^{\prime}(t)=k y+b\) in the case that \(k y+b<0\) and verify that the general solution is \(y(t)=C e^{k t}-\frac{b}{k}\)

Sociologists model the spread of rumors using logistic equations. The key assumption is that at any given time, a fraction \(y\) of the population, where \(0 \leq y \leq 1,\) knows the rumor, while the remaining fraction \(1-y\) does not. Furthermore, the rumor spreads by interactions between those who know the rumor and those who do not. The number of such interactions is proportional to \(y(1-y) .\) Therefore, the equation that describes the spread of the rumor is \(y^{\prime}(t)=k y(1-y)\) where \(k\) is a positive real number. The number of people who initially know the rumor is \(y(0)=y_{0},\) where \(0 \leq y_{0} \leq 1\) a. Solve this initial value problem and give the solution in terms of \(k\) and \(y_{0}\) b. Assume \(k=0.3\) weeks \(^{-1}\) and graph the solution for \(y_{0}=0.1\) and \(y_{0}=0.7\) c. Describe and interpret the long-term behavior of the rumor function, for any \(0 \leq y_{0} \leq 1\)

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. a. If the growth rate function for a population model is positive, then the population is increasing. b. The solution of a stirred tank initial value problem always approaches a constant as \(t \rightarrow \infty\) c. In the predator-prey models discussed in this section, if the initial predator population is zero and the initial prey population is positive, then the prey population increases without bound.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free