Taking derivatives of DE to find higher derivatives

  • Thread starter shemer77
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  • #1
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Homework Statement


http://gyazo.com/d875970963124b7d4a64acc887f168fa

The Attempt at a Solution


I dont understand the part where they take the derivative of a diff eq to find the higher derivatives can somebody explain it?
Mainly the answer where they get
y''=(y'')'=(1-y^2)'=-2yy'(this part)
y^(4)=(y''')'=(-2yy')'=-2y'y'-2yy"(huh?)=-2(y')^2-2yy''
 

Answers and Replies

  • #2
tiny-tim
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hi shemer77! :smile:

(try using the X2 button just above the Reply box :wink:)
y'''=(y'')'=(1-y2)'=-2yy'(this part)

since you have an equation for y'', you use it to simplify all the higher derivatives

(1-y2)' = (-y2)' = (use the chain rule) (-2y)(y)'
y(4)=(y''')'=(-2yy')'=-2y'y'-2yy"(huh?)=-2(y')^2-2yy''

similarly, use the product rule for (-2yy')' :wink:
 
  • #3
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Hmm ok I think I understand the second one but for the first one what I've been taught is
(1-y2)'=1'-y2'=-2y
where does that extra y' come from?
 
  • #4
tiny-tim
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… for the first one what I've been taught is
(1-y2)'=1'-y2'=-2y
where does that extra y' come from?

remember that ' means d/dx, not d/dy

so the chain rule gives you

y2' = d/dx (y2)

= d/dy (y2) * dy/dx

= 2y * y' :wink:
 

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