# 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.