Desperately need help for a 2nd order diff eq!

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Kim679

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Hello: I need to explain the solution to a simple 2nd order differential equation tomorrow for an engineering course (I'm a TA) and I've totally forgotten how to solve differential equations. I have the general solution to the following equation, but I can't remember the steps. Any help would be greatly appreciated!

(d^2h/dr^2)+1/r(dh/dr)=0

Thank you so much!
 
Kim679 said:
Hello: I need to explain the solution to a simple 2nd order differential equation tomorrow for an engineering course (I'm a TA) and I've totally forgotten how to solve differential equations. I have the general solution to the following equation, but I can't remember the steps. Any help would be greatly appreciated!

(d^2h/dr^2)+1/r(dh/dr)=0

Thank you so much!
Click to expand...
Can you slove:

df/dr + f/r = 0
 
Not sure...I forget (I was on maternity leave for a year and am having trouble remembering everything).
ln(f)+ln(r)+C=0?
 
Kim679 said:
Not sure...I forget (I was on maternity leave for a year and am having trouble remembering everything).
ln(f)+ln(r)+C=0?
Click to expand...
I know it is very hard to be a Mom and go to school>

But you need to dust-up the books and the notes and let's go...

ln(f)+ln(r)+C=0 ..................... CORRECT ................... however I would write it as:

ln(f) + ln(K*r)=0...............where C = ln(K) .....................................edited

ln(f) = - ln(K*r) = ln(C1/r)....... where C1 = 1/K............edited

f = C1/r

Now how is it related to original problem?

We have replaced f (= dh/dr) to get the DE that you had solved.

So:

dh/dr = C1/r

Now solve for 'h'......
 
Last edited by a moderator:
Thanks for your help. I think the answer is:
h = C1 ln r +C2
(I understand how to solve dh/dr=C1/r...but the first step with the 2nd order differential is still confusing me (even with your helpful explanation)
 
Subhotosh Khan said:
ln(f)+ln(r)+C=0 ..................... CORRECT ................... however I would write it as:

ln(f) + ln(C*r)=0
Click to expand...
There's a "oops" moment here. Let C = ln(c). Then
ln(f) + ln(r) + C = 0

ln(f) + ln(r) + ln(c) = 0

ln(f) + ln(rc) = 0

-Dan
 
Kim679 said:
Thanks for your help. I think the answer is:
h = C1 ln r +C2
(I understand how to solve dh/dr=C1/r...but the first step with the 2nd order differential is still confusing me (even with your helpful explanation)
Click to expand...
What is that step?
 
Thanks to you both for your help. Unfortunately, I'm still confused. Thanks anyway!
 
Kim679 said:
Thanks to you both for your help. Unfortunately, I'm still confused. Thanks anyway!
Click to expand...
Kim,

Exactly where are you getting lost? Please point it out.

It will be rather painless - you have done it before. Since you are going to be TA - you need to have all your doubts clarified to be successful. Tell us exactly where you are stuck....
 
Thanks for continuing to help, Subhotosh!
I know I need to do two integrations to solve for h. I am stuck on how to write out the very first steps. This is what I was thinking I need to do:

Step 1:
integrate (d2h/dr2) dr = integrate (-1/r )(dh/dr) dr

dh/dr = integrate (-1/r)dh

dh/dr + C = -h/r +C


(I think I'm going wrong at with the right hand side?)

Should it be

dh/dr + C = -1/r +C

(That is, I'm stuck on the first integration. Once it gets down to dh/dr = C1/r, and the second integration is needed, I am ok.)

Thank you!
 
Let's kind of start over:

(d^2h/dr^2) + 1/r(dh/dr) = 0 ...... you cannot directly integrate this - so we substitute (assume):

f = dh/dr

Then the equation above becomes:

df/dr + f/r = 0 .............................. (response #2)..........................this is a first order equation. It "can" be integrated directly.

\(\displaystyle \int\) df/f = -\(\displaystyle \int\) dr/r

ln(f) = - ln(r) + ln(C1) ...................... constant of integration.

ln(f) = ln(C1/r)

f = C1/r

f = dh/dr = C1/r

Clear upto here ..... any doubts .....

Integrate again and continue to solve for 'h'
 
Thank you very much! This helps a lot!!! I didn't realize the substitution was required.

The only part I'm a bit confused about now is:


f = dh/dr

Then the equation above becomes:

df/dr + f/r = 0

-> why is the df/dr written in that way? That is, it is now (dh/dr)/dr (if you were to substitute back in for the f). Is that the same meaning as the 2nd derivitaive of h with respect to r?

I'm sorry if this is a silly question!
 
Kim679 said:
Thank you very much! This helps a lot!!! I didn't realize the substitution was required.

The only part I'm a bit confused about now is:


f = dh/dr

Then the equation above becomes:

df/dr + f/r = 0

-> why is the df/dr written in that way? That is, it is now (dh/dr)/dr (if you were to substitute back in for the f). Is that the same meaning as the 2nd derivitaive of h with respect to r?

I'm sorry if this is a silly question!
Click to expand...
This is not a silly question. But does mean that you need to review differential equations from the begining.

df/dr = d/dr(f) = d/dr(dh/dr) = d2h/dr2
 
I would choose to multiply the original ODE by \(r\)

[MATH]r\frac{d^2h}{dr^2}+\frac{dh}{dr}=0[/MATH]
Observe, this may now be written as:

[MATH]\frac{d}{dr}\left(r\frac{dh}{dr}\right)=0[/MATH]
Integrating with respect to \(r\), we obtain:

[MATH]r\frac{dh}{dr}=c_1[/MATH]
Or:

[MATH]\frac{dh}{dr}=\frac{c_1}{r}[/MATH]
Integrating again with respect to \(r\), we obtain:

[MATH]h(r)=c_1\ln|r|+c_2[/MATH]
 
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