Help for solving a 2nd order non-linear ODE

  • Thread starter AdrianZ
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  • #1
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Homework Statement


[tex]yy''-y'^2 = y^2lny[/tex]

The Attempt at a Solution


well, since the equation is of the form [itex]f(y,y',y'')=0[/itex] I turn it into the form [itex]f(y,p,p dp/dy)=0[/itex].
After those substitutions are made, we'll have the following equation:
[tex]yp (\frac{dp}{dy})-p^2-y^2 lny=0[/tex]
which is a Bernoulli equation that can be solved easily. the solution of this ODE is:
[tex]p^2=y^2lny+cy^2[/tex]
Here's where I'm stuck, because If I substitute y'=p again I'll have an ODE that I don't know how to solve it, I guess the general solution will be parametric, but I don't know how to proceed from this step.
 
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Answers and Replies

  • #2
Stephen Tashi
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[tex]yy''-y'^2 = y^2lny[/tex]
One thought is:
[tex] D( \frac{y'}{y}) = ln(y) [/tex]
[tex] \frac{y'}{y} = \int ln(y) [/tex]
 
  • #4
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One thought is:
[tex] D( \frac{y'}{y}) = ln(y) [/tex]
[tex] \frac{y'}{y} = \int ln(y) [/tex]
Would you explain more please?

Have you tried a substitution y(x)=eq(x) ?
I don't know how that would work. Would you explain more?
 
  • #5
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One thought is:
[tex] D( \frac{y'}{y}) = ln(y) [/tex]
[tex] \frac{y'}{y} = \int ln(y) [/tex]
Would you explain more please?
Stephen has these backwards.

D(ln(y)) = y'/y
so ln(y) = ∫(y'/y)
 
  • #6
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Stephen has these backwards.

D(ln(y)) = y'/y
so ln(y) = ∫(y'/y)
Yes, I understood that, but how does that help?

I think this is what he means:
[itex]y'^2 = y^2 lny + cy^2[/itex]→ [itex]y^{-2}y'^2-lny=C[/itex]
→ [itex]\frac{y'^2}{y^2} - lny = C[/itex]→ [itex](\frac{y'}{y})^2 - (\int{\frac{y'}{y}dy})=C[/itex]
Hence, the equation has turned into the form [itex]u'^2 - u = C[/itex]. Has anything changed? Can I solve this one?
 
  • #7
Stephen Tashi
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Stephen has these backwards.

D(ln(y)) = y'/y
so ln(y) = ∫(y'/y)
That's not my thought. My thought is that the original equation is equivalent to the equation [itex] D(\frac{y'}{y}) = ln(y) [/itex]

But employing what you said:

[tex] D(\frac{y'}{y}) = \int \frac{y'}{y} [/tex]
 
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  • #8
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Ahh, now I see what you mean. That's a great way of doing this ODE once you know that D(y'/y)=(yy''-y'^2)/y^2. Is there anyway to proceed with my own method?
 
  • #9
Stephen Tashi
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Is there anyway to proceed with my own method?
I don't see how to use your method. As understand the situation, the solution of the Bernoulli equation that you employ assumes that [itex] p [/itex] is a known function. But [itex] p [/itex] isn't known.
 
  • #10
dextercioby
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I don't know how that would work. Would you explain more?
It would get you rid of the natural logarithm and perhaps the transformed ODE would be integrable...
 
  • #11
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the solution of this ODE is:
[tex]p^2=y^2lny+cy^2[/tex]
Your solution is wrong at this step!

The correct solution is:

[tex]
p y^2 (C + \ln^2{y}) = 1, \ p = y'
[/tex]

If you split the variables, the integral over y is expressible in terms of elementary functions.
 

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