fp2, some ****ed up differentiation question Watch

anus eater
Badges: 0
Rep:
?
#1
Report Thread starter 7 years ago
#1
y(d^2y/dx^2) + (dy/dx)^2 + y = 0

Find an expression for d^3y/dx^3

differentiating (product rule for y(d^2y/dx^2)):
(d^2y/dx)(dy/dx) + y(d^3y/dx^3) + d^3y/dx^3 + y(dy/dx) = 0

typing that alone required more effort than trying to sneeze, masturbate and take a **** at the same time (and i'm guessing i've already gone wrong too according to the mark scheme), so just put me out of my misery . what did i do wrong?
0
reply
nuodai
Badges: 14
#2
Report 7 years ago
#2
You differentiated \left( \dfrac{dy}{dx} \right)^2 and y incorrectly. For the first you can use the chain rule, and the second is trivial (because \dfrac{d}{dx}(y) = \dfrac{dy}{dx}).

With these things I prefer to write it using dash notation; that is:

yy'' + (y')^2 + y = 0

Then when you differentiate anything you just increase the number of dashes; so for example, when you differentiate yy'' you get y'y'' + yy''' (this is the part you did correctly). You just need to differentiate (y')^2 and y and then rearrange the resulting equation to make y''' the subject.
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • Cranfield University
    Cranfield Forensic MSc Programme Open Day Postgraduate
    Thu, 25 Apr '19
  • University of the Arts London
    Open day: MA Footwear and MA Fashion Artefact Postgraduate
    Thu, 25 Apr '19
  • Cardiff Metropolitan University
    Undergraduate Open Day - Llandaff Campus Undergraduate
    Sat, 27 Apr '19

Have you registered to vote?

Yes! (285)
37.75%
No - but I will (53)
7.02%
No - I don't want to (55)
7.28%
No - I can't vote (<18, not in UK, etc) (362)
47.95%

Watched Threads

View All