Help for solving a 2nd order non-linear ODE

1. Nov 16, 2011

1. The problem statement, all variables and given/known data
$$yy''-y'^2 = y^2lny$$

3. The attempt at a solution
well, since the equation is of the form $f(y,y',y'')=0$ I turn it into the form $f(y,p,p dp/dy)=0$.
After those substitutions are made, we'll have the following equation:
$$yp (\frac{dp}{dy})-p^2-y^2 lny=0$$
which is a Bernoulli equation that can be solved easily. the solution of this ODE is:
$$p^2=y^2lny+cy^2$$
Here's where I'm stuck, because If I substitute y'=p again I'll have an ODE that I don't know how to solve it, I guess the general solution will be parametric, but I don't know how to proceed from this step.

Last edited: Nov 16, 2011
2. Nov 16, 2011

Stephen Tashi

One thought is:
$$D( \frac{y'}{y}) = ln(y)$$
$$\frac{y'}{y} = \int ln(y)$$

3. Nov 16, 2011

dextercioby

Have you tried a substitution y(x)=eq(x) ?

4. Nov 16, 2011

I don't know how that would work. Would you explain more?

5. Nov 16, 2011

Staff: Mentor

Stephen has these backwards.

D(ln(y)) = y'/y
so ln(y) = ∫(y'/y)

6. Nov 16, 2011

Yes, I understood that, but how does that help?

I think this is what he means:
$y'^2 = y^2 lny + cy^2$→ $y^{-2}y'^2-lny=C$
→ $\frac{y'^2}{y^2} - lny = C$→ $(\frac{y'}{y})^2 - (\int{\frac{y'}{y}dy})=C$
Hence, the equation has turned into the form $u'^2 - u = C$. Has anything changed? Can I solve this one?

7. Nov 16, 2011

Stephen Tashi

That's not my thought. My thought is that the original equation is equivalent to the equation $D(\frac{y'}{y}) = ln(y)$

But employing what you said:

$$D(\frac{y'}{y}) = \int \frac{y'}{y}$$

Last edited: Nov 16, 2011
8. Nov 16, 2011

Ahh, now I see what you mean. That's a great way of doing this ODE once you know that D(y'/y)=(yy''-y'^2)/y^2. Is there anyway to proceed with my own method?

9. Nov 17, 2011

Stephen Tashi

I don't see how to use your method. As understand the situation, the solution of the Bernoulli equation that you employ assumes that $p$ is a known function. But $p$ isn't known.

10. Nov 17, 2011

dextercioby

It would get you rid of the natural logarithm and perhaps the transformed ODE would be integrable...

11. Nov 17, 2011

Dickfore

Your solution is wrong at this step!

The correct solution is:

$$p y^2 (C + \ln^2{y}) = 1, \ p = y'$$

If you split the variables, the integral over y is expressible in terms of elementary functions.