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UnkleRhaukus

  • 5 years ago

\[y{d^2y \over dx^2}+1=({dy \over dx})^2\]

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  1. UnkleRhaukus
    • 5 years ago
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    the equation is x-absent

  2. anonymous
    • 5 years ago
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    y y'' +1= (y')^2 (y'y)'=y'' y+ (y')^2

  3. UnkleRhaukus
    • 5 years ago
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    i can not see how you did that

  4. anonymous
    • 5 years ago
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    that's not quite right since y'' y+ (y')^2 and we want y'' y- (y')^2

  5. UnkleRhaukus
    • 5 years ago
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    let\[y'=p\]\[y''=pp'\]

  6. anonymous
    • 5 years ago
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    can you please retype your equation it's confusing me

  7. anonymous
    • 5 years ago
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    can you please retype your equation it's confusing me

  8. anonymous
    • 5 years ago
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    can you please retype your equation it's confusing me

  9. anonymous
    • 5 years ago
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    can you please retype your equation it's confusing me

  10. anonymous
    • 5 years ago
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    can you please retype your equation it's confusing me

  11. anonymous
    • 5 years ago
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    can you please retype your equation it's confusing me

  12. anonymous
    • 5 years ago
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    can you please retype your equation it's confusing me

  13. UnkleRhaukus
    • 5 years ago
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    ok

  14. UnkleRhaukus
    • 5 years ago
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    \[y=y(x)\]\[y{d^2y \over dx^2}+1=({dy \over dx})^2\]or\[y{y''}+1=(y')^2\]

  15. UnkleRhaukus
    • 5 years ago
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    \[ypp'+1=p^2\]

  16. UnkleRhaukus
    • 5 years ago
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    \[p'+{1 \over py} = {p \over y}\]

  17. JamesJ
    • 5 years ago
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    Differentiate your equation wrt x and we have \[ y'y'' + yy''' = 2y'y'' \] i.e., \[ yy''' = y'y'' \] or \[ \frac{y'''}{y''} = \frac{y'}{y} \] Now integrate wrt to find \[ \ln(y'') = \ln(y) + C \] or \[ y'' - Ay = 0 \] for a positive constant \( A (= -e^C) \). The solutions of this are now easy to find and you can see they satisfy the original ODE. Notice that if you not take A to be positive, you wind end with sin and cos as a basis for the solution space, and those functions do not satisfy the original ODE.

  18. JamesJ
    • 5 years ago
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    *correction: A = +e^C

  19. UnkleRhaukus
    • 5 years ago
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    the answer in the back of my book is \[±cln(y+\sqrt{y^2+c^2} = x +d\] where c and d are the constants

  20. JamesJ
    • 5 years ago
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    The thing on the left looks like arccosh, so that makes sense to me. I'm trying to figure out though where they get two constants from.

  21. UnkleRhaukus
    • 5 years ago
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    well we have y''=... so integration twice will produce the two constants

  22. UnkleRhaukus
    • 5 years ago
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    right?

  23. JamesJ
    • 5 years ago
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    Yes, I know in general a second order _linear_ equation has a 2-dimensional solution space. And having two parameters here is also consistent. I'm just trying to figure out where I lost one it in the solution.

  24. JamesJ
    • 5 years ago
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    Ok. So writing more carefully, we have \[ y'' - \lambda^2 y = 0 \] and two possible Ansatz are \[ y_1(x) = A \cosh(\lambda (x-x_0)) \] \[ y_2(x) = B \sinh(\lambda (x - x_0)) \] Now, it's not hard to show y_2 can't satisfy the original equation, and y_1 satisfies it if and only \[ A^2 \lambda^2 \cosh^2(\lambda(x-x_0)) + 1 = A^2 \lambda^2 \cosh^2(\lambda(x-x_0)^2) \] i.e., \[ A = 1/\lambda \] Hence it appears one 2-parameter family of solutions is \[ y_{(\lambda,x_0)}(x) = \cosh(\lambda(x-x_0)) \]

  25. JamesJ
    • 5 years ago
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    I know this procedure is somewhat unsatisfactory and feels a bit arbitrary. Part of that, I think, is the nature of non-linear ODEs. I suppose we can call on the Picard theorem to convince ourselves there aren't more solutions.

  26. JamesJ
    • 5 years ago
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    bleah ... that equation 2 posts above should stuff . cosh^2 + 1 = stuff . sinh^2

  27. UnkleRhaukus
    • 5 years ago
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    i got lost

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