A community for students.
Here's the question you clicked on:
 0 viewing
UnkleRhaukus
 4 years ago
\[y{d^2y \over dx^2}+1=({dy \over dx})^2\]
UnkleRhaukus
 4 years ago
\[y{d^2y \over dx^2}+1=({dy \over dx})^2\]

This Question is Closed

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1the equation is xabsent

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0y y'' +1= (y')^2 (y'y)'=y'' y+ (y')^2

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1i can not see how you did that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's not quite right since y'' y+ (y')^2 and we want y'' y (y')^2

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1let\[y'=p\]\[y''=pp'\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you please retype your equation it's confusing me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you please retype your equation it's confusing me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you please retype your equation it's confusing me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you please retype your equation it's confusing me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you please retype your equation it's confusing me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you please retype your equation it's confusing me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0can you please retype your equation it's confusing me

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1\[y=y(x)\]\[y{d^2y \over dx^2}+1=({dy \over dx})^2\]or\[y{y''}+1=(y')^2\]

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1\[p'+{1 \over py} = {p \over y}\]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Differentiate 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.

UnkleRhaukus
 4 years ago
Best ResponseYou've already chosen the best response.1the 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
 4 years ago
Best ResponseYou've already chosen the best response.2The 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
 4 years ago
Best ResponseYou've already chosen the best response.1well we have y''=... so integration twice will produce the two constants

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2Yes, I know in general a second order _linear_ equation has a 2dimensional 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
 4 years ago
Best ResponseYou've already chosen the best response.2Ok. So writing more carefully, we have \[ y''  \lambda^2 y = 0 \] and two possible Ansatz are \[ y_1(x) = A \cosh(\lambda (xx_0)) \] \[ y_2(x) = B \sinh(\lambda (x  x_0)) \] Now, it's not hard to show y_2 can't satisfy the original equation, and y_1 satisfies it if and only \[ A^2 \lambda^2 \cosh^2(\lambda(xx_0)) + 1 = A^2 \lambda^2 \cosh^2(\lambda(xx_0)^2) \] i.e., \[ A = 1/\lambda \] Hence it appears one 2parameter family of solutions is \[ y_{(\lambda,x_0)}(x) = \cosh(\lambda(xx_0)) \]

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2I know this procedure is somewhat unsatisfactory and feels a bit arbitrary. Part of that, I think, is the nature of nonlinear ODEs. I suppose we can call on the Picard theorem to convince ourselves there aren't more solutions.

JamesJ
 4 years ago
Best ResponseYou've already chosen the best response.2bleah ... that equation 2 posts above should stuff . cosh^2 + 1 = stuff . sinh^2
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.