2nd order ODE reduction of order method

  • Thread starter pat666
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  • #1
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Homework Statement



[tex] x^2 y''+2xy'-12y=0,y_1=x^3 [/tex]
[tex] y''+(2y')/x-12y/x^2 =0 [/tex]

Homework Equations





The Attempt at a Solution


[tex] y=uy_1=ux y'=u' x+u y''=u'' x+2u' [/tex]
subbing that in instead of y,y',y''
[tex] u'' x+2u'+2(u'x+u)/x+12(ux)/x^2 =0 [/tex]

now my book says to reduce the order v=u' which is easy enough except for the fact that I have "u" in there with no derivative. What do I sub here (integral of v)????? STUMPED!

thanks for any help
 

Answers and Replies

  • #2
lurflurf
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x is not a solution, x^3 is.
 
  • #3
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Yes, thats one mistake so far:
[tex] y_1=uy_1=u*x^3 y'=u'*x^3+u y''=u''x^3+2u' [/tex]
before going on can you tell me if thats correct??

thanks
 
  • #4
lurflurf
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I do not follow what you wrote, the subscripts are unclear and it all runs together.
The product rule is
(uv)'=u'v+uv'
(uv)''=u''v+2u'v'+uv''
 
  • #5
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sorry, Ill fix it:
[tex] y_1=u*x^3--------------y'=u'*x^3+u----------y''=u''x^3+2u' [/tex]
 
  • #6
lurflurf
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(x^3)'=3 x^2
(x^3)''=6 x
You might be differentiating x
x'=1
x''=0
 
  • #7
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sorry, I don't know what you mean. I was following how the textbook does it which doesn't seem to be normal differentiation????
 
  • #8
lurflurf
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It is normal differentiation.
let
y=u y1
y'=(uy1)'=u'y1+u y1'
y''=(uy1)''=u''y1+2u'y1'+uy1''
in this case
y1=x^3
y1'=3x^2
y1''=6x
so
y=x^3 u
y'=x^3 u'+3x^2 u
y''=x^3 u''+6x^2 u'+6x u
which we substitute into the differential equation
 
  • #9
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Again your right, I was treating u as a constant and not what it actually is.
so what I have now is:
[tex] x^3*u''+6x^2u'+6x*u+2(x^3 u'+3x^2 u)/x-12(x^3 u)/x^2=0 [/tex]
now I can let v=u' or do I neaten things up first?
 
  • #10
lurflurf
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You can let v=u' if you want, you will notice the u's cancel out as expected.
 
  • #11
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Hi again, just neatened it up, in actually comes out quite neatly:
[tex] x^3u''+8x^2u' =0[/tex]
then subbing v=u', separation of variables + integration gives
[tex] ln(v)=-8ln(x) [/tex] can you give me some more help here please.
 
  • #12
lurflurf
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Just reverse the substitutions
v=A x^-8
u'=v
y=x^3 u
to find y
y will have the form A y1+B y2
where A and B are arbitrary constants
 
  • #13
HallsofIvy
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Hi again, just neatened it up, in actually comes out quite neatly:
[tex] x^3u''+8x^2u' =0[/tex]
then subbing v=u', separation of variables + integration gives
[tex] ln(v)=-8ln(x) [/tex] can you give me some more help here please.
Well, [itex]-8ln(x)= ln(x^{-8})[/itex], of course, so you have [itex]ln(v)= ln(x^{-8})[/itex]. Can you solve that for v? And then integrate to find u.
 
  • #14
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Certainty can:
[tex] v=1/x^8 [/tex]
since v=u' so ' = integral of v
so [tex]u=\int 1/x^8 [/tex]
[tex]u=-1/(7x^7) [/tex]
Do I need a constant of integration here or not because I am looking for a particular solution?

Thanks

also not sure what the next step is?
 
  • #15
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Hey got a little further:
I have y=x^3u so u=y/x^3
and u=-A/(7x^7 )
so y/x^3=-A/(7x^7 )
But I don't have A and B??

Thanks
 
  • #16
HallsofIvy
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Actually there are two constants of integration and you (and I) completely forgot the first one- which is the one you need.

ln(v)= -8ln(x)+ c gives [itex]v= Cx^{-8}[/itex] ([itex]C= e^c[/itex].)

Then [itex]u'= v= Cx^{-8}[/itex] so [itex]u(x)= -\frac{C}{7}x^{-7}+ C_2[/itex]

And then [itex]y(x)= u(x)x^3= -\frac{C}{7}x^{-4}+ C_2x^3[/itex] is the general solution. Notice that the second constant of integration gives the original solution, [itex]x^3[/itex]. Of course, you could have ignored both constants of integration to just get that [itex]x^{-4}[/itex] is a second independent solution and, already knowing that [itex]x^3[/itex]is a solution, write the general solution as
[tex]y(x)= Ax^{-4}+ Bx^3[/tex].
 
  • #17
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Thanks heaps for all your help HallsofIvy and lurflurf! mucho appreciato!
 

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