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Second Order Nonlinear ODE

  • Thread starter josephsuk
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  • #1
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1. y''y^4 = 8



I tried almost every method I know, including laplace transforms, variation of parameters, reductin of order, v=y' substitution
 

Answers and Replies

  • #2
1,796
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First, you realize that y'' can be written as d(y') and y' as dy right? So what happens if you multiply both sides of the equation by y'?

Well, you get:

[tex]y'd(y')y^4=8dy[/tex]

Now what?
 
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  • #3
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It appears to be logical to use separation of variables to get y'd(y')=8y^-4dy
and then integrate to get the solution?
 
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  • #4
1,796
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It appears to be logical to use separation of variables to get [tex]y'd(y')=8 dy[/tex]
Forgot the other part. I get:

[tex]y'd(y')=8y^{-4} dy[/tex]
 
  • #5
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I am not exactly what follows in the next steps.

Would integrating y' d(y') give me ((y')^2)/2 or something else?

If it does give me that answer, then I get y to be [tex]y=\frac{5}{2}x\sqrt[5]{\frac{-16}{3}}[/tex]
 
  • #6
LCKurtz
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I am not exactly what follows in the next steps.

Would integrating y' d(y') give me ((y')^2)/2 or something else?

If it does give me that answer, then I get y to be [tex]y=\frac{5}{2}x\sqrt[5]{\frac{-16}{3}}[/tex]
Look at your given DE and see if you think a constant times x can be a solution.
 
  • #7
1,796
53
I am not exactly what follows in the next steps.

Would integrating y' d(y') give me ((y')^2)/2 or something else?

If it does give me that answer, then I get y to be [tex]y=\frac{5}{2}x\sqrt[5]{\frac{-16}{3}}[/tex]
No way dude. you got:

[tex]\int y' d(y')=\int 8 y^{-4} dy[/tex]
and that's
[tex]\frac{(y')^2}{2}=-8/3 y^{-3}+c[/tex]
Ok, can you now separate variables (take square root first), and then post what the next integral expression would be? Can't integrate it directly (not easily) but just the expression of what has to be integrated is good enough for now.
 

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