2nd order nonlinear DE need help bad

  • Thread starter j_reez
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  • #1
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this is the problem: xy'' -x(y')^2 = y'

my book says that i need to substitute u=y' and du/dx=y''...
so i get:

x(du/dx)-xu^2 = u

so next the book says i need to separate the x's/dx' to one side and u's/du's to the other. however, i cannot do it

am i using the correct technique?


thanks-
Justin
 

Answers and Replies

  • #2
Galileo
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edit: Nevermind
 
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  • #3
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this problems driving me crazy :surprised
 
  • #4
ehild
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j_reez said:
this is the problem: xy'' -x(y')^2 = y'

my book says that i need to substitute u=y'


Substitute xy'=u instead.

ehild
 
  • #5
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i dont see how that would work....if u=xy' then du/dx=y' +xy''
how would thos substitutions work?
 
  • #6
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if you want you can work it like this:

x(du/dx)-xu^2=u => (u^-2)(du/dx)-(u^-1)/x=1 then use this

substitution, I think this is "Bernoulli equation" z=-(u^-1) and dz/dx=(u^-2)(du/dx)

=> dz/dx +z/x=1. I hope this help you.
 
  • #7
James R
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Your equation is:

[tex]xy'' -x(y')^2 = y'[/tex]

Since y does not appear in the equation, start by putting:

[tex]y' = u[/tex]

which means the equation is:

[tex]xu' - xu^2 = u[/tex]
[tex]u' - u^2 = \frac{u}{x}[/tex]

Now put

[tex]u(x) = xv(x)[/tex]

So that

[tex]u' = v'x + v[/tex]

The differential equation then becomes separable, as follows:

[tex]v'x + v - v^2 x^2 = v[/tex]
[tex]xv' = v^2 x^2[/tex]
[tex]dv/v^2 = x dx[/tex]

Integrating gives:

[tex]-\frac{1}{v} = \frac{x^2}{2} + c[/tex]

where c is an undetermined constant. Rewriting v in terms of u gives:

[tex]-\frac{x}{u} = \frac{x^2}{2} + c[/tex]

Rearranging, we get:

[tex]-\frac{u}{x} = \frac{1}{x^2/2 + c}[/tex]

or

[tex]y' = -\frac{x}{x^2/2 + c}[/tex]

This can now be easily integrated to give the solution:

[tex]y = -\ln(x^2/2 + c) + d[/tex]

where d is a second constant of integration.
 
  • #8
ehild
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j_reez said:
i dont see how that would work....if u=xy' then du/dx=y' +xy''
how would thos substitutions work?

[tex]y' = \frac{u}{x}\mbox { , } y''=\frac{u'x-u}{x^2}[/tex]

Plugging in y' and y'' the differential equation becomes

[tex]\frac{xu'-u}{x}-\frac{u^2}{x}=\frac{u}{x}\rightarrow u'x=u^2+2u[/tex]

This is easily separable:

[tex]\frac{du}{u^2+2u}=\frac{dx}{x}[/tex]

and results in

[tex]u=\frac{2x^2}{C-x^2} \mbox { that is } y'=\frac{2x}{C-x^2}[/tex]

This is the same solution James got but his method is simpler, I have to admit :smile: You see there are many ways to go....

ehild
 
  • #9
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Thank You

thanks guys! its very clear to me now :) my book had one very limited example which made it sound like i only need to make one substitution. however, i clearly see it now!

thanks,
Justin
 

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