Solving a Differential Equation: Y''*Y^2=C with Real Number C

[gerechte23]

Hi I've been striving to find the solution of this differential equation, but can't find: Y''*Y^2=C c being a real number. Please give me a hand with it. Thanks in advance!

 

[phyzguy]

Differential equations like this, which do not contain the independent variable, are called autonomous equations. Try:

http://en.wikipedia.org/wiki/Autonomous_system_(mathematics )

 

[gerechte23]

Thanks for the source. I'll make my best to solve it. But i would like to have a person to help me with it though. I barely can use the internet. I am in Haiti.

Can u try helping me with it please!

 

[epenguin]

You can get your first integral fairly easily. For this type of d.e. where x does not appear, you can define p = Y'. Then

Y'' = \frac{dp}{dx} = \frac{dp}{dY}.\frac{dY}{dx} = p\frac{dp}{dY}

so you have a first-order equation in the variables p and Y.

You can integrate that but then you have to do a second integration. Except if you take the constant of integration zero you have a rather nasty square root I think, but it can be done; if you don't know how maybe someone will help if you come back with the first part.

 

[gerechte23]

hi buddy,
Thanks for your help. Actually, it's the very same method a man here has proposed to me. And i ended up having one equation where i have my Y on one side, and the variable x on the other.
When i kept integrating it gives me something with logarithm and square root with Y. i wanted to find Y in respect of x, but what i find is X in respect of y. i don't know how to calculate it so i find Y in respect of x. It's so weird and difficult to me cause i don't know any method that can help.
Later today i will post the result here so that you give me an idea. I don't have the papers with me now.

Thanks again, i thought i would find another method for it. i really appreciate your help

 

[phyzguy]

It sounds like you're on the right track. After you get x in terms of y, you need to invert this relation to get y in terms of x. Post what you have so we can see.

 

[gerechte23]

Here is the equation it gave me. I resolved it, i didn't expect to come to the internet so i didn't come with the papers, so is it right? \int\sqrt{\frac{y}{by-c}} dy = \int\sqrt{2} dx. b and c are real numbers!

 

[epenguin]

Here is the equation it gave me. I resolved it, i didn't expect to come to the internet so i didn't come with the papers, so is it right? \int\sqrt{\frac{y}{by-c}} dy = \int\sqrt{2} dx. b and c are real numbers!

I think so, at least I got the same result except I made a mistake you didn't. So on the right you have got x basically. (NB c is your given constant whereas b is an arbitrary constant.)

As I said you have to do a second integration - it is not clear to me whether you have, but you mentioned weird results which make it sound that you have. If you come back with more it shall be revealed to you, unless you twigged it already, how they are not weird as they may seem but quite to be expected and useful!