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
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."
How do you get that? I'm pretty sure that the first integral is a fairly straightforward computation once you make a substitution.
[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/m_{0}) then you end up with the integral of 1/u with respect to u.
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.
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/m_{0}) the integral will turn out very nicely.
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, thats 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).
[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.
Yes but then you have another integral to solve via more substitution. Nothing wrong with your method, just longer.
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.
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.