How to solve an indefinite integral of a differential equation?

In summary, the conversation discusses the process of solving a differential equation involving an indefinite integral. The participants suggest making a substitution, u = log(m/mo), to simplify the integral and arrive at the solution. They also mention the importance of choosing the correct u, using the ILATE method.
  • #1
NEWO
95
0
I have a formula to solve, which I am having trouble solving, it is a differential equation fo sorts

[tex]\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}[/tex]

could someone help me solve the indefinite integral of such an equation!

the m multiply of the right handside is suppose to be dm and likewise with the t on the left hand side sorry for any confusion!
Thanks for any help

N
 
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  • #2
NEWO said:
I have a formula to solve, which I am having trouble solving, it is a differential equation fo sorts

[tex]\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}[/tex]

could someone help me solve the indefinite integral of such an equation!

the m multiply of the right handside is suppose to be dm and likewise with the t on the left hand side sorry for any confusion!
Thanks for any help

N

Make a substitution u=log(m/mo). I assumed that mo in the denominator was supposed to be mo.
 
  • #3
How can this work? Its an indefinite integral and the answer's definate. One's with respect to m and the other t. I'm sure it doesn't change the u substitution, but shouldn't everything be noted as rigorously as possible?
 
  • #4
Dbjergaard said:
How can this work? Its an indefinite integral and the answer's definate. One's with respect to m and the other t. I'm sure it doesn't change the u substitution, but shouldn't everything be noted as rigorously as possible?

I think everything is as already as rigorous as possible, at least for all that the original poster has given, however, I have no idea what you mean by "Its an indefinite integral and the answer's definate."
 
  • #5
I guess that was wrong.
 
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  • #6
robert Ihnot said:
Something like [tex]\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}[/tex] the first part which could be written as [tex]\int{\frac{e^zdz}{z}}[/tex] I don't suppose has a definite integral in closed form.

How do you get that? I'm pretty sure that the first integral is a fairly straightforward computation once you make a substitution.
 
  • #7
[tex]\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}[/tex]

Let m=e^x then we get [tex]\int{\frac{dx}{x-log m(0)}[/tex]
 
  • #8
robert Ihnot said:
[tex]\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}[/tex]

Let m=e^x then we get [tex]\int{\frac{dx}{x-log m(0)}[/tex]

You're missing a factor of e-x which then gives you something you can't integrate, but if you make the substitution u=log(m/m0) then you end up with the integral of 1/u with respect to u.
 
  • #9
We may just notice that 1/m is the 1st derivative of (log(m/m0)) and that, then, the first integral contains an expression dividing its derivative: integrate[D(log(m/m0))]/log(m/m0)]dm. But this is just log|log(m/m0)|+c; isn't it?
 
  • #10
MWM said:
We may just notice that 1/m is the 1st derivative of (log(m/m0)) and that, then, the first integral contains an expression dividing its derivative: integrate[D(log(m/m0))]/log(m/m0)]dm. But this is just log|log(m/m0)|+c; isn't it?

Yes that's it, and this is the same solution you arrive at using the substitution I mentioned, and is pretty much the sam process.
 
  • #11
Ah, you're right. Ok... as if I hadn't said anything :-)
 
  • #12
Ok thanks guys for your replies,

the thing is I don't get where the m has gone!

surely the integral would be

[tex]\int{\frac{1}{m_{0}ue^{u}}}du[/tex]

??
 
  • #13
NEWO said:
Ok thanks guys for your replies,

the thing is I don't get where the m has gone!

surely the integral would be

[tex]\int{\frac{1}{m_{0}ue^{u}}}du[/tex]

??

How do you get to that? Try using the substitution u=log(m/m0) the integral will turn out very nicely.
 
  • #14
Let [itex]u=\log(m/mo)[/itex]
The integral:[tex]\int{\frac{1}{m\log(m/mo)}{dm}[/tex] becomes [tex]\int{\frac{1}{m\cdot u}{dm}[/tex]. Whats the derivative of log x? 1/x of course.
[tex]u=\log (m/mo)[/tex]
[tex]\frac {du}{dm} =\frac{1}{m}[/tex]
Substituting that in, our integral becomes [tex]\int \frac{1}{u} \frac {du}{dm} dm[/tex].
Dm's cancel out, that's why your m goes away!
And obviously [tex]\int \frac{1}{u} du = \log u [/tex].

Now that looks the same form as the right hand side of your first post. Therefore [itex]t=\log (m/mo)[/itex]. Hopefully that was what you are looking for.
 
  • #15
Gib Z said:
Let [itex]u=\log(m/mo)[/itex]
The integral:[tex]\int{\frac{1}{m\log(m/mo)}{dm}[/tex] becomes [tex]\int{\frac{1}{m\cdot u}{dm}[/tex]. Whats the derivative of log x? 1/x of course.
[tex]u=\log (m/mo)[/tex]
[tex]\frac {du}{dm} =\frac{1}{m}[/tex]
Substituting that in, our integral becomes [tex]\int \frac{1}{u} \frac {du}{dm} dm[/tex].
Dm's cancel out, that's why your m goes away!
And obviously [tex]\int \frac{1}{u} du = \log u [/tex].

Now that looks the same form as the right hand side of your first post. Therefore [itex]t=\log (m/mo)[/itex]. Hopefully that was what you are looking for.

You are right but I want to add,
ln(m/mo)=t+c, c is the constant of integration
Simplifying, ln(m)=t+k, k= c+ ln(mo).
 
  • #16
O yes sorry, forgot about the Constant, my bad. Ty for correcting me.
 
  • #17
[tex]\int{\frac{1}{m\log(m/mo)}{dm}=\int{\frac{1}{t}}{dt}[/tex]

What was wrong with substiting X=Log m?

[tex]\int{\frac{dx}{x-log m(0)}[/tex]

Remember, if e^x=m then, e^xdx =dm; so ms in the top and bottem cancel out.
 
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  • #18
Yes but then you have another integral to solve via more substitution. Nothing wrong with your method, just longer.
 
  • #19
My natural instinct was to substitute u= ln(m/mo) , for the probable shortest way, I saw later 'Gib Z' also did the same substitution.
 
  • #20
Theres a special order to choose your u. Its very helpful: LIATE. Logs, Inverse Trig, Algebraic, Trig, and Exponentials. Go down the list and pick the first one as your u. In that case its obvious this time its the Log.
 
  • #21
Gib Z said:
Theres a special order to choose your u. Its very helpful: LIATE. Logs, Inverse Trig, Algebraic, Trig, and Exponentials. Go down the list and pick the first one as your u. In that case its obvious this time its the Log.

Probably you mean ILATE, for choosing the 1st function for integration by parts.
 
  • #22
O no, I am 100% sure i ment LIATE, I've seen ILATE, but I was taught LIATE and i haven't had any problems, everyman to his own :)
 
  • #23
Since L and I are have higher preference than others, the preference between L & I does not matter if you have a function containing one of L or I. But when only L & I functions are there, I is first and then comes L... this is what my class notes of 11th std said. I just have checked a textbook after your post and that confirms mine. I really do not know what the truth is... but had a bias for ILATE as the textbooks mentioned ... and never questioned it before today.
 
  • #24
Gib Z said:
Let [itex]u=\log(m/mo)[/itex]
The integral:[tex]\int{\frac{1}{m\log(m/mo)}{dm}[/tex] becomes [tex]\int{\frac{1}{m\cdot u}{dm}[/tex]. Whats the derivative of log x? 1/x of course.
[tex]u=\log (m/mo)[/tex]
[tex]\frac {du}{dm} =\frac{1}{m}[/tex]
Substituting that in, our integral becomes [tex]\int \frac{1}{u} \frac {du}{dm} dm[/tex].
Dm's cancel out, that's why your m goes away!
And obviously [tex]\int \frac{1}{u} du = \log u [/tex].

Now that looks the same form as the right hand side of your first post. Therefore [itex]t=\log (m/mo)[/itex]. Hopefully that was what you are looking for.

Correct.
How do you get the special math text to appear?
Cheers,
 
  • #25
You have to learn laTex code. Theres a tutorial somehwere, I can't remember...I didnt learn from the tutorial btw, I clicked on other peoples code and learned from that. Click on my code to see what I typed to get that to appear.
 
  • #26

What is the integral of 1/ln(x) formula?

The integral of 1/ln(x) formula is a mathematical expression that represents the area under the curve of the function 1/ln(x). It is a commonly used formula in calculus and is used to solve problems involving logarithmic functions.

How is the integral of 1/ln(x) formula derived?

The integral of 1/ln(x) formula is derived using integration by substitution. This involves substituting u = ln(x) and du = dx/x into the integral and then solving for u. The resulting integral can then be solved using known integration techniques.

What is the domain of the integral of 1/ln(x) formula?

The domain of the integral of 1/ln(x) formula is all positive real numbers, excluding 1. This is because the natural logarithm function, ln(x), is only defined for positive numbers and the denominator of 1/ln(x) cannot equal 0.

Can the integral of 1/ln(x) formula be simplified?

Yes, the integral of 1/ln(x) formula can be simplified by using the properties of logarithms and integrating by parts. The result will be an expression involving the natural logarithm function and the original function, 1/ln(x).

What are the practical applications of the integral of 1/ln(x) formula?

The integral of 1/ln(x) formula has practical applications in various fields such as physics, economics, and engineering. It is commonly used to solve problems related to exponential growth and decay, as well as in calculating the area under logarithmic curves in real-world scenarios.

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