# Help for solving a 2nd order non-linear ODE

## Homework Statement

$$yy''-y'^2 = y^2lny$$

## 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.

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## Answers and Replies

Stephen Tashi
$$yy''-y'^2 = y^2lny$$
One thought is:
$$D( \frac{y'}{y}) = ln(y)$$
$$\frac{y'}{y} = \int ln(y)$$

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

One thought is:
$$D( \frac{y'}{y}) = ln(y)$$
$$\frac{y'}{y} = \int ln(y)$$
Would you explain more please?

Have you tried a substitution y(x)=eq(x) ?
I don't know how that would work. Would you explain more?

Mark44
Mentor
One thought is:
$$D( \frac{y'}{y}) = ln(y)$$
$$\frac{y'}{y} = \int ln(y)$$
Would you explain more please?
Stephen has these backwards.

D(ln(y)) = y'/y
so ln(y) = ∫(y'/y)

Stephen has these backwards.

D(ln(y)) = y'/y
so ln(y) = ∫(y'/y)
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?

Stephen Tashi
Stephen has these backwards.

D(ln(y)) = y'/y
so ln(y) = ∫(y'/y)
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:

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

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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?

Stephen Tashi
Is there anyway to proceed with my own method?
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.

dextercioby
Homework Helper
[...]
I don't know how that would work. Would you explain more?
It would get you rid of the natural logarithm and perhaps the transformed ODE would be integrable...

the solution of this ODE is:
$$p^2=y^2lny+cy^2$$
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.