UnkleRhaukus
  • UnkleRhaukus
\[y{d^2y \over dx^2}+1=({dy \over dx})^2\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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UnkleRhaukus
  • UnkleRhaukus
the equation is x-absent
anonymous
  • anonymous
y y'' +1= (y')^2 (y'y)'=y'' y+ (y')^2
UnkleRhaukus
  • UnkleRhaukus
i can not see how you did that

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anonymous
  • anonymous
that's not quite right since y'' y+ (y')^2 and we want y'' y- (y')^2
UnkleRhaukus
  • UnkleRhaukus
let\[y'=p\]\[y''=pp'\]
anonymous
  • anonymous
can you please retype your equation it's confusing me
anonymous
  • anonymous
can you please retype your equation it's confusing me
anonymous
  • anonymous
can you please retype your equation it's confusing me
anonymous
  • anonymous
can you please retype your equation it's confusing me
anonymous
  • anonymous
can you please retype your equation it's confusing me
anonymous
  • anonymous
can you please retype your equation it's confusing me
anonymous
  • anonymous
can you please retype your equation it's confusing me
UnkleRhaukus
  • UnkleRhaukus
ok
UnkleRhaukus
  • UnkleRhaukus
\[y=y(x)\]\[y{d^2y \over dx^2}+1=({dy \over dx})^2\]or\[y{y''}+1=(y')^2\]
UnkleRhaukus
  • UnkleRhaukus
\[ypp'+1=p^2\]
UnkleRhaukus
  • UnkleRhaukus
\[p'+{1 \over py} = {p \over y}\]
JamesJ
  • JamesJ
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.
JamesJ
  • JamesJ
*correction: A = +e^C
UnkleRhaukus
  • UnkleRhaukus
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
JamesJ
  • JamesJ
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.
UnkleRhaukus
  • UnkleRhaukus
well we have y''=... so integration twice will produce the two constants
UnkleRhaukus
  • UnkleRhaukus
right?
JamesJ
  • JamesJ
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.
JamesJ
  • JamesJ
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 \[ y_{(\lambda,x_0)}(x) = \cosh(\lambda(x-x_0)) \]
JamesJ
  • JamesJ
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.
JamesJ
  • JamesJ
bleah ... that equation 2 posts above should stuff . cosh^2 + 1 = stuff . sinh^2
UnkleRhaukus
  • UnkleRhaukus
i got lost

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