## UnkleRhaukus 5 years ago $y{d^2y \over dx^2}+1=({dy \over dx})^2$

1. UnkleRhaukus

the equation is x-absent

2. anonymous

y y'' +1= (y')^2 (y'y)'=y'' y+ (y')^2

3. UnkleRhaukus

i can not see how you did that

4. anonymous

that's not quite right since y'' y+ (y')^2 and we want y'' y- (y')^2

5. UnkleRhaukus

let$y'=p$$y''=pp'$

6. anonymous

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12. anonymous

13. UnkleRhaukus

ok

14. UnkleRhaukus

$y=y(x)$$y{d^2y \over dx^2}+1=({dy \over dx})^2$or$y{y''}+1=(y')^2$

15. UnkleRhaukus

$ypp'+1=p^2$

16. UnkleRhaukus

$p'+{1 \over py} = {p \over y}$

17. 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.

18. JamesJ

*correction: A = +e^C

19. 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

20. 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.

21. UnkleRhaukus

well we have y''=... so integration twice will produce the two constants

22. UnkleRhaukus

right?

23. 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.

24. 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 $y_{(\lambda,x_0)}(x) = \cosh(\lambda(x-x_0))$

25. 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.

26. JamesJ

bleah ... that equation 2 posts above should stuff . cosh^2 + 1 = stuff . sinh^2

27. UnkleRhaukus

i got lost