# Solving the second order equation?

• Feb 25th 2013, 06:48 PM
colerelm1
Solving the second order equation?
How do I solve an equation that is missing the independent variable?

$2y^2* y'' + 2y(y')^2 = 1$
• Feb 25th 2013, 07:58 PM
jakncoke
Re: Solving the second order equation?
I have an idea but i'm not sure if it will work

If y = $ax^2 + bx + c$ then
$2y^2*y''$ = 4th degree polynomial
$2y*(y')^2 =$ 4th degree polynomial

so you would have to solve for the coefficients so that they cancel out.
• Feb 26th 2013, 12:10 AM
JJacquelin
Re: Solving the second order equation?
Quote:

Originally Posted by jakncoke
I have an idea but i'm not sure if it will work

If y = $ax^2 + bx + c$ then
$2y^2*y''$ = 4th degree polynomial
$2y*(y')^2 =$ 4th degree polynomial

so you would have to solve for the coefficients so that they cancel out.

Hi !

I don't think that the method proposed by jakncoke is convenient. OK., the order of the polynomials are the same. But this doesn't mean that they can be equal. In some case yes, but not in the present case.
The solving is more difficult (see in attachment)
• Feb 26th 2013, 03:25 AM
HallsofIvy
Re: Solving the second order equation?
I should have read JJacqueline's post first- she did exactly the same thing.
• Feb 26th 2013, 04:29 AM
JJacquelin
Re: Solving the second order equation?
Quote:

Originally Posted by HallsofIvy
I should have read JJacqueline's post first- she did exactly the same thing.

"Clothes do not make the man" : No "e" at the end of the name of the man ! (Giggle)
Thanks for the thanks
JJ.