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

Show that (7.15) is a separable equation. [You may find it helpful to write.] F(x)dx=f(x)Thus solve (7.14) in terms of quadrature’s (that is, indicated integrations) as in Problem 2.

Short Answer

Expert verified

The solution of the differential equation is ±dx2m[f(x)+A]=t+const

Step by step solution

01

Given information from question

The given equation is 12mv2=F(x)dx+const.

02

Velocity

The differential form of velocity is v(x)=dxdt.

03

Calculate the solution of the differential equation

The given equation is separable:

12mv2=F(x)dx+const

The integral on the RHS should be carried out for some variable other thanx , but up tox,along these lines:

0xF(y)dy

The above integral as a function ofx

f(x)=0xF(y)dy

Now transform the initial equation into:

12mv2(x)=f(x)+A

Divide the equation bym2to obtain an expression forv(x):

v2(x)=2m[f(x)+A]v(x)=±2m[f(x)+A)

04

Apply definition of velocity as a time derivative of position v(x)=dxdt

The definition of velocity as a time derivative of positionv(x)=dxdtand inserting it into the previous expression, so that transformation the initial condition into:

dxdt=±2m[f(x)+A]±dx2mf(x)+A=dt

At last, integrate the previous equation and use the table integraldx=x+const on theRHS

to write the general form of its situation:

±dx2m[f(x)+A]=t+const

Thus, the solution of the differential equation is data-custom-editor="chemistry" ±dx2m[f(x)+A]=t+constconst.

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

See all solutions

Recommended explanations on Physics 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