# Non linear 2nd order differential equation

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• Jul 25th 2009, 07:30 AM
Leo Klem
Non linear 2nd order differential equation
25 July 2009

I am not able to find the general function that integrates the following 2nd order differential equation;

y^2 y’’ + a y^3b = 0

in which 0 ≤ a ≤ 1 is a constant;
b > 0 is a constant.

The equation relates to the behaviour (orbit/motion) of a massless point subjected to a particular central motion.
I could easily find only the solutions associated with nil values for constants a and b.

I would be deeply grateful to any friend who can help me resolve the problem.
Many thanks for any help.
Leo Klem
• Jul 25th 2009, 08:15 AM
Deco
Uhm, let me see if I know what you're asking.

$y'' = -ay + \frac{b}{y^2}$

$y' = -a \int {y} dy + b \int {y^{-2}} dy$

$y' = -a(\frac {y^2}{2} + C_1) + b(-y^{-1} + C_2)$

$y' = -a\frac {y^2}{2} - by^{-1} + C_t$

$y = -a \int (\frac {y^2}{2})dy - b \int (y^{-1})dy + \int C_t dy$

$y = -a (\frac {y^3}{6} + C_3) - b (\ln y + C_4) + C_ty + C_5$

$y = - \frac {ay^3}{6} - b \ln y + C_ty + C_T$

It's either that or I'm totally off the mark. (Rock)
• Jul 25th 2009, 01:38 PM
Jester
Quote:

Originally Posted by Deco
Uhm, let me see if I know what you're asking.

$y'' = -ay + \frac{b}{y^2}$

$y' = -a \int {y} dy + b \int {y^{-2}} dy$

$y' = -a(\frac {y^2}{2} + C_1) + b(-y^{-1} + C_2)$

$y' = -a\frac {y^2}{2} - by^{-1} + C_t$

$y = -a \int (\frac {y^2}{2})dy - b \int (y^{-1})dy + \int C_t dy$

$y = -a (\frac {y^3}{6} + C_3) - b (\ln y + C_4) + C_ty + C_5$

$y = - \frac {ay^3}{6} - b \ln y + C_ty + C_T$

It's either that or I'm totally off the mark. (Rock)

You're off the mark on both integrations (but close on the first)!

If you multiply your equation by $y'$ then

$y'y'' = -ay y' + \frac{b y'}{y^2}$

so

$
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1
$

You can solve for $y'$ and separate but I think the resulting integral will be difficult.
• Jul 25th 2009, 01:51 PM
Deco
$y' = \sqrt {- a y^2} - \sqrt {\frac{2b}{y}} + c_1$

$y'= \sqrt {-a} \int y dy - \sqrt {2b} \int y^{-\frac {1}{2}} dy + \int C_1 dy$

$y = \sqrt {-a} ( \frac {y^2}{2}) - \sqrt {2b} (2y^{\frac {1}{2}}) + C_1y + C_2$

Something like that?

But since a is positive, wouldn't this result in a complex answer?
• Jul 25th 2009, 01:56 PM
Jester
Quote:

Originally Posted by Deco
$y' = \sqrt {- a y^2} - \frac{2b}{y} + c_1$

$y'= \sqrt {-a} \int y dy - \sqrt {2b} \int y^{-\frac {1}{2}} dy + \int C_1 dy$

$y = \sqrt {-a} ( \frac {y^2}{2}) - \sqrt {2b} (2y^{\frac {1}{2}}) + C_1y + C_2$

Something like that?

How on earth did you go from

$
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1$

to

$y' = \sqrt {- a y^2} - \frac{2b}{y} + c_1$?
• Jul 25th 2009, 01:58 PM
Deco
Oops. Wow. Don't know what I was thinking.
• Jul 26th 2009, 03:54 AM
Leo Klem
Non linear 2nd order equation.Thanks, with a clarification
Many thanks to my two first interlocutors. I have to point out that the problem described by the differential equation regards a central motion. The integrating function, if it does exist, is actually useful in the explicit form given by
y = f(x; a, b, C, D)

in which x is the independent variable (it represent the "orbital" angle), a and b are the given equation constants, and C, D are the two integration constants.
Solutions in the form of an expression of functions of y do actually create difficult problems of interpretation as to the kind of motion regarded.
• Jul 26th 2009, 11:17 PM
chisigma
Quote:

Originally Posted by Danny
How on earth did you go from

$
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1$

to

$y' = \sqrt {- a y^2} - \frac{2b}{y} + c_1$?

It is evident that the correct procedure is...

$
\frac{y'^2}{2} = - \frac{a y^2}{2} - \frac{b}{y} + c_1 \rightarrow$

$\rightarrow y^{'}= +/- \sqrt{\frac{- a y^{3} - 2 b + 2 y c_{1}}{y}} \rightarrow$

$\rightarrow \int \sqrt{\frac{y}{-a y^{3} -2b + 2 y c_{1}}} dy = +/- x + c_{2}$ (1)

May be that the integral in (1) is a little difficult to solve ... may be (Itwasntme)...

Kind regards

$\chi$ $\sigma$
• Jul 27th 2009, 02:37 AM
Leo Klem
What beyond the inverse function?
Thanks to Chisigma. You got the correct procedure, but the solution is still away. It's the point where I had actually to stop. It's clearly a "solution" in the form of an inverse function, i.e., the argument expressed as a function of the dependent variable. The integral you indicate seems however difficult to solve, but in a few particular cases, such as - for instance - assuming C1 = 0.
Could you please provide further suggestions?
Nome a parte, sono italiano anch'io, ma non sono un matematico. Grato comunque.(Hi)