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Confused what method to use on this Diff EQ! 2nd order

  • Thread starter mr_coffee
  • Start date
  • #1
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ello ello!
I havn't seen an example of this type of problem. It has another variabler in the equation, i tried to divide by it, but its still in there. Here is the problem:
Find y as a function of x if
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/ee/448e92b4900790fd8de8a4207ce40d1.png [Broken]
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/c9/0bc438e10b7ec37bb81942786780e41.png [Broken]
y = ?
So i divided by x^2 and got:
y''+2y'/x -30y/x^2 = x;

This kind of looks like an integrating factor problem but not. ANy suggestions or can someone explain to me whats going on? Thanks! :biggrin:
 
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Answers and Replies

  • #2
I've not looked at these type of equations before either but i'll speculate an answer if it's alright with the sire.

You're told the values of y' and y for when x is 1 so substitute them in with x as 1 to get:

(1^2)y'' + 2(1x5) - (30x6) = 1
y'' = 171

Then integrate to get: y' = 171x + c

but you're told that this is equal to 5 when x is 1 so that...

5 = 171 + c
c = -166
therefore y' = 171x - 166

then integrate again and find the constant again and you should get that..

y = 171(x^2)/2 -166x + 173/2

Don't hold me to that, i'm just frying logic
 
  • #3
p.s. i'm new here. Where is everyone getting their maths type?
 
  • #4
1,629
1
Rooftop, its LaTex.
And i subbmited that answer and it wasn't right, i have infinately many tries so i'm not worried about submitting wrong answers. Anyone else know?
 
  • #5
thanks for the heads up...

btw... you wouldn't happen to be doing some kind of radical calculus crossword puzzle would you?

edit: wait..... i don't seem to be finding any LaTex. I may need a bigger heads up...
 
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  • #6
HallsofIvy
Science Advisor
Homework Helper
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RooftopDuvet: y"= 171 only when x= 1. y'= 171x+ C only if y"= 171 for all x.

mr_coffee: You are asking the same kind of questions about the same kinds of problems over and over: This is an "Euler type" equation (also called "equipotential" equation). Remember what I told you about them last time?
 
  • #7
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Sorry Ivey, i post too much! For it to be a Euler Type doesn't i have to "be centerd around" a value?
like http://tutorial.math.lamar.edu/AllBrowsers/3401/EulerEquations_files/eq0001M.gif [Broken]
around Xo = 0. These type of differential equations are called Euler Equations. But mine doesn't loook like that because its not homogenous, and it doens't say its centered around anything, do i have to modify it or somthing?

I've solved Euler Equations beofre like this one:
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/26/a0c710e74525576a0593cd3a5bf7881.png [Broken]
and got it right, but it was equated to 0, not x
 
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  • #8
335
4
mr_coffee said:
ello ello!
I havn't seen an example of this type of problem. It has another variabler in the equation, i tried to divide by it, but its still in there. Here is the problem:
Find y as a function of x if
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/ee/448e92b4900790fd8de8a4207ce40d1.png [Broken]
http://cwcsrv11.cwc.psu.edu/webwork2_files/tmp/equations/c9/0bc438e10b7ec37bb81942786780e41.png [Broken]
y = ?
So i divided by x^2 and got:
y''+2y'/x -30y/x^2 = x;

This kind of looks like an integrating factor problem but not. ANy suggestions or can someone explain to me whats going on? Thanks! :biggrin:
Worst comes to worst you can always use my favorite method: Use the substitution u = ln x. That reduces the problem to a 2nd order linear with constant coefficients.

-Dan
 
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  • #9
1,629
1
topsquark, what are u subbing? u = ln x? where do u put that?
 
  • #10
HallsofIvy
Science Advisor
Homework Helper
41,792
920
An "Euler type equation" has x to a power equal to the order of the derivative (that's why it's also called "equipotential"). Your example,
x2y"+ 2xy'- 30y= x3 is exactly of that kind. As for the substitution, u= ln x, I've also told you about that before.

If u= ln x, by the chain rule, dy/dx= (du/dx)(dy/du)= (1/x)dy/du and
d2y/dx2= (-1/x2)dy/du+ (1/x2)d2y/du2. Also x= eu so x3= e3u Your equation becomes
d2y/du2+ dy/du- 30u= e3u.
That should be easy.
 

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