Integrate y''/y: Solving with Logs

[*FaerieLight*]

Homework Statement

How would you find the integral of y''/y (with respect to x)?

Homework Equations

The Attempt at a Solution

I have absolutely no idea how to begin. All I know is that if it were y'/y, it would be log(y) + c. Perhaps the integral here has something to do with logs.

Thanks a lot.

 

[Tomer]

Try integration in parts, where f(x) = y''(x), g'(x) = 1/y(x).
You should arrive to a new integral you can solve...

 

[HallsofIvy]

Tomer's suggestion is good but I think a more common notation would be u= 1/y, dv= y'' dy.

 

[LCKurtz]

Homework Statement

How would you find the integral of y''/y (with respect to x)?

Homework Equations

The Attempt at a Solution

I have absolutely no idea how to begin. All I know is that if it were y'/y, it would be log(y) + c. Perhaps the integral here has something to do with logs.

Thanks a lot.

Try integration in parts, where f(x) = y''(x), g'(x) = 1/y(x).
You should arrive to a new integral you can solve...

Tomer's suggestion is good but I think a more common notation would be u= 1/y, dv= y'' dy.

I don't see how either of these substitutions help to solve this problem. y is a function of x and the variable of integration is x.

 

[Tomer]

I don't see how either of these substitutions help to solve this problem. y is a function of x and the variable of integration is x.

I haven't for a second assumed x isn't the variable of integration.
However, trying to solve it now on paper I noticed that what I first though would be a solvable integral isn't really.
Do you have any ideas then?

 

[jackmell]

If:

\frac{d}{dx}\left[\frac{y'}{y}+\left(\frac{d}{dx}\log y\right)^2\right]=\frac{y''}{y}

then isn't:

\int \frac{y''}{y}=\frac{y'}{y}+\left(\frac{d}{dx}\log y\right)^2

Not entirely sure guys. Just a start.

 

[Tomer]

If:

\frac{d}{dx}\left[\frac{y'}{y}+\left(\frac{d}{dx}\log y\right)^2\right]=\frac{y''}{y}

then isn't:

\int \frac{y''}{y}=\frac{y'}{y}+\left(\frac{d}{dx}\log y\right)^2

Not entirely sure guys. Just a start.

What you say is right, but I don't see how the first formula you wrote is correct.

 

[jackmell]

What you say is right, but I don't see how the first formula you wrote is correct.

Ok, it's wrong. Sorry.

 

[Tomer]

It's ok, apparently we're all wrong :-)

 

[PAllen]

It's ok, apparently we're all wrong :-)

No, it seems that the starting point in post #6 can't possibly be right. It could only be right if the derivative of the second term in brackets is zero, which can't possibly be true, in general.

 

[LCKurtz]

I will be a bit more emphatic. You aren't going to find a nice closed form general solution with any such techniques.

 

[PAllen]

I will be a bit more emphatic. You aren't going to find a nice closed form general solution with any such techniques.

Yeah, it is equivalent to asking what is a general formula for the solution of the following homogeneous, linear second order equation with non-constant coefficients:

y'' - f(x) y = 0

which is absurd (unless f(x) is special in some way).

 

[Tomer]

Yeah, it is equivalent to asking what is a general formula for the solution of the following homogeneous, linear second order equation with non-constant coefficients:

y'' - f(x) y = 0

which is absurd (unless f(x) is special in some way).

Why is this absurd? If I can say that \int\frac{y'(x)}{y(x)} = ln(y(x)), which is a closed form general solution, why should the given integral be absurd?

I don't see any way to find a closed formula, but I also don't see why the question should be meaningless.

 

[PAllen]

Why is this absurd? If I can say that \int\frac{y'(x)}{y(x)} = ln(y(x)), which is a closed form general solution (and is equivalent to finding f(x) in the equation y'(x) +f(x)y(x) = 0), why should the given integral be absurd?

I don't see any way to find a closed formula, but I also don't see why the question should be meaningless.

It isn't meaningless, it is just well understood - it has been studied (stated as diff.eq.) for centuries. All facts about it are known. It is known that there are solutions for many common f(x), but no formula for a general solution.

 

[Tomer]

It isn't meaningless, it is just well understood - it has been studied (stated as diff.eq.) for centuries. All facts about it are known. It is known that there are solutions for many common f(x), but no formula for a general solution.

I understand :p