# Thread: Separation in differential equation

1. ## Separation in differential equation

How can I get a separate y, y' and y'' in equation

(y')^2 + y*y''+y'*y = 0

Best regards

2. I'm not entirely sure I understand your question. However, I would note that

$\dfrac{d}{dx}(y^{2})=2yy',$ and hence

$\dfrac{d^{2}}{dx^{2}}(y^{2})=2[(y')^{2}+yy''],$ and hence

$\dfrac{1}{2}\,\dfrac{d^{2}}{dx^{2}}(y^{2})=(y')^{2 }+yy''.$

The DE becomes

$\dfrac{(y^{2})''}{2}+yy'=0.$ We can now integrate directly, once, to obtain

$\dfrac{1}{2}\,(y^{2})'+\dfrac{y^{2}}{2}=C_{1},$ or re-defining $C_{1},$ we have

$(y^{2})'+y^{2}=C_{1}.$

Can you continue?

3. Thank you Mr Ackbeet. I can continue. Anyway, I want to separate y,y' and y'' in diff equation to get in form ay''+by'+cy+g(y'^2). How can we do this if we can? Thank a lot.

4. Hmm. That might be difficult. The substitution $u=y^{2}$ yields the DE

$u''+u'=0,$

which is technically in the form you're looking for, with $a=b=1,$ and $c=g=0.$

The substitution $u=yy'$ gives you the same DE, actually, just one order lower:

$u'+u=0.$

In this case, you have $a=g=0,$ and $b=c=1.$

I don't know if that helps you or not. That's a strange form to try to get the DE into - at least I'm not familiar with it. It looks semi-Ricatti, but in a second-order form. I don't know.

Simply dividing through by $yy'$ yields the immediately integrable expression

$\dfrac{y'}{y}+\dfrac{y''}{y'}+1=0.$

Other than this, I'm about out of ideas.