help solving a 2nd order differential equation

[dmw16]

I have the equation:

y'' = -1/y

does anyone know how to reduce it to a first order? I can't remember. I thought it was something like let u = y' and then sub, but then I get intergral of u on the left side...
-doug

 

[wasssup]

isn't the derivative of ln (x) = 1/x?

in this case, i think the first derivative would be:

-ln y + C (where C is a constant)

 

[Flyermax2k3]

Originally posted by: dmw16
I have the equation:

y'' = -1/y

does anyone know how to reduce it to a first order? I can't remember. I thought it was something like let u = y' and then sub, but then I get intergral of u on the left side...
-doug

Is that supposed to be y squared or a 2nd iteration of y? If it's squared I would think you would just take the square root of -1/y but what do I know?

 

[kt]

Originally posted by: Flyermax2k3

Originally posted by: dmw16
I have the equation:

y'' = -1/y

does anyone know how to reduce it to a first order? I can't remember. I thought it was something like let u = y' and then sub, but then I get intergral of u on the left side...
-doug

Is that supposed to be y squared or a 2nd iteration of y? If it's squared I would think you would just take the square root of -1/y but what do I know?

He did say "2nd order differential equation", so go figure. Besides, y squared would be denoted as y^2.

 

[dmw16]

it has nothing to do w/ squares or square roots...

 

[esun]

I'm sure I have this in my notes somewhere, I just have to look it up. Check later if no one else comes up with the solution.

EDIT: Well I found it online: http://www.sosmath.com/diffeq/second/nonlineareq/nonlineareq.html

Check the section for the condition that x is missing.

In short:

v = dy/dx
d2y/dx2 = vdv/dy

Make those two substitutions.

 

[hdeck]

it's been too long (a whole semester) since i took diff equations or i'd help you =\

 

[tigerbait]

Originally posted by: hdeck
it's been too long (a whole semester) since i took diff equations or i'd help you =\

I took it three years ago, and I got a headache from reading this thread 😕

 

[Flyermax2k3]

Originally posted by: kt

Originally posted by: Flyermax2k3

Originally posted by: dmw16
I have the equation:

y'' = -1/y

does anyone know how to reduce it to a first order? I can't remember. I thought it was something like let u = y' and then sub, but then I get intergral of u on the left side...
-doug

Is that supposed to be y squared or a 2nd iteration of y? If it's squared I would think you would just take the square root of -1/y but what do I know?

He did say "2nd order differential equation", so go figure. Besides, y squared would be denoted as y^2.

Well there ya go! Sorry, I haven't had a math course in 3 years, it's hard to remember these things if you don't use them.

 

[RaynorWolfcastle]

do you have any initial conditions for the equation?

actually, esun's solution is right... I was thinking of using Laplace