Question

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

Answer 1

the equation is x-absent

Answer 2

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

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

Answer 3

i can not see how you did that

Answer 4

that's not quite right since
y'' y+ (y')^2

and we want

y'' y- (y')^2

Answer 5

let\[y'=p\]\[y''=pp'\]

Answer 6

can you please retype your equation it's confusing me

Answer 7

can you please retype your equation it's confusing me

Answer 8

can you please retype your equation it's confusing me

Answer 9

can you please retype your equation it's confusing me

Answer 10

can you please retype your equation it's confusing me

Answer 11

can you please retype your equation it's confusing me

Answer 12

can you please retype your equation it's confusing me

Answer 13

ok

Answer 14

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

Answer 15

\[ypp'+1=p^2\]

Answer 16

\[p'+{1 \over py} = {p \over y}\]

Answer 17

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.

Answer 18

*correction: A = +e^C

Answer 19

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

Answer 20

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.

Answer 21

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

Answer 22

right?

Answer 23

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.

Answer 24

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
\[ y_{(\lambda,x_0)}(x) = \cosh(\lambda(x-x_0)) \]

Answer 25

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.

Answer 26

bleah ... that equation 2 posts above should

stuff . cosh^2 + 1 = stuff . sinh^2

Answer 27

i got lost