Results are out! Find what you need...fast. Get quick advice or join the chat
x

Unlock these great extras with your FREE membership

  • One-on-one advice about results day and Clearing
  • Free access to our personal statement wizard
  • Customise TSR to suit how you want to use it

Simple Differential Equation - Help Me Please

Announcements Posted on
Find your uni forum to get talking to other applicants, existing students and your future course-mates 27-07-2015
  1. Offline

    ReputationRep:
    Question:

    Given that  \displaystyle y = u^{-1}

    And  \displaystyle \dfrac{du}{dx} - ux^{-1} = -x^{-1}\ln x

    And  \displaystyle y(1)=\frac12

    Find  \displaystyle y=f(x)

    My pdf textbook is notorious for getting things wrong so I would appreciate it if someone could verify my answer, I won't post my full working but:

    Integrating factor:  \displaystyle I = x^{-1}

    Hence  \displaystyle ux^{-1} = \int -x^{-2}\ln x \ dx

    After using integration by parts on the RHS:  \displaystyle ux^{-1} = -x^{-1}\ln x - x^{-1} +C

    Now I get  \displaystyle u = -\ln x -1 + C

    And the book gets  \displaystyle u = \ln x +1 +C

    Without solving for y, which is just the reciprocal, can anyone confimr who is correct ?

    Thnx
  2. Offline

    ReputationRep:
    the book is correct.

    the integration by parts is:

    Integral[ {-1/x^2}{Log[x]} = - (-1/x)(Log[x]) + Integral[(-1/x)(1/x)]dx

    = 1/x+(log[x])/x +C =u/x
  3. Offline

    ReputationRep:
    Sorry, but you are both wrong!

    The correct answer is,

    u = ln(x) - 1 + Cx

    In your IBP, you should have ux^{-1}=+x^{-1}ln(x)-x^{-1}+C

    If you work out \frac{du}{dx}-ux^{-1}=-x^{-1}ln(x)

    using my answer, you will see that it satisfies the DE.
  4. Offline

    ReputationRep:
    (Original post by steve10)
    Sorry, but you are both wrong!

    The correct answer is,

    u = ln(x) - 1 + Cx

    In your IBP, you should have ux^{-1}=+x^{-1}ln(x)-x^{-1}+C

    If you work out \frac{du}{dx}-ux^{-1}=-x^{-1}ln(x)

    using my answer, you will see that it satisfies the DE.
    My mistake was that I took the -1 out of the bracket and forgot to put it back in,
    The correct answer is  u = \ln x +1 +cx .
  5. Offline

    ReputationRep:
    Some really wrong answers on here.. I agree with most of what you've said and Steve10 is the closest.

    The correct answer would be u=ln(x)+x+cx
  6. Offline

    ReputationRep:
    (Original post by member910132)
    My mistake was that I took the -1 out of the bracket and forgot to put it back in,
    The correct answer is  u = \ln x +1 +cx .
    I can confirm this is the correct answer
  7. Offline

    ReputationRep:
    my answer, equated to u/x , then multiplied through by x, is correct.

    (k = 1) and y=1/(x+Log(x)+1)

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: May 17, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
Have you ever posted a picture of your food on social media?
Results and Clearing

Results are coming...

No sweat. Here's all you need to make sure you're ready

new on tsr

Join the American Society

Whether you're from the USA or just love it!

Study resources
x

Think you'll be in clearing or adjustment?

Hear direct from unis that want to talk to you

Get email alerts for university course places that match your subjects and grades. Just let us know what you're studying.

Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.