how to solve this?

logistic_guy said:
\(\displaystyle \frac{dy}{dx}\ln(\frac{dy}{dx}) = c\)
Click to expand...
You have developed a habit of simply posting questions that you want answers to.

You must submit your own efforts to answer a question when you post it so that it may be seen where you are having difficulty and the best advice to aid you in overcoming that difficulty may then be offered.

This forum is here to help you to do your work, not to do your work for you.

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as explained here:-


Please now share your work/thoughts about the above problems.
 
the structure is new. if i seperate \(\displaystyle dy \ln(dy) - \ln(dx) = c dx\) i dont think this is valid. the only thing i can do random guesses
 
i think i figure the answer \(\displaystyle y(x) = c_1x\), \(\displaystyle \frac{dy}{dx} = c_1\), \(\displaystyle c_1\ln(c_1) = c\), but it say wrong answer
 
i made random guess, i choose \(\displaystyle y(x) = c_1\), \(\displaystyle \frac{dy}{dx} = 0\), \(\displaystyle 0\ln(0) = c\) i think no valide so i choose \(\displaystyle y(x) = c_1x\)
 
logistic_guy said:
\(\displaystyle \frac{dy}{dx}\ln(\frac{dy}{dx}) = c\)
Click to expand...
This is a differential equation - yet you posted in "arithmetic" forum!

What topic are being taught in your class-room (real or virtual)?

Is this a home-work problem? If it is - post the complete problem!
 
It's differential course.

this is the question

Find the solution, \(\displaystyle y(x)\), of the differential equation \(\displaystyle \frac{dy}{dx}\ln(\frac{dy}{dx}) = c\)
can you please tell me why my answer wrong?
 
\(\displaystyle u = \frac{dy}{dx}\),
\(\displaystyle u\ln u = c\),
\(\displaystyle e^u u = e^c = c_1\),
\(\displaystyle u = c_1e^{-u}\)
may be wrong calculation i don't know

how my post 6 my answer is incomplete?
if \(\displaystyle u = \frac{dy}{dx} = c_1\), then \(\displaystyle c_1\ln(c_1) = c\), i see it true
 
This differential equation can have the famous form [imath]z = xe^x[/imath] which can be solved in terms of the Lambert W function [imath]W(z) = x[/imath]. In fact [imath]W(z)[/imath] is the inverse of [imath]z[/imath].
 
Last edited by a moderator:
blamocur said:
That would be my answer, yes.
Click to expand...
it say it correct, but it want me to express the answer in terms of the constant

for example if the differential equation is \(\displaystyle \frac{dy}{dx}\ln(\frac{dy}{dx}) = 5\), it want \(\displaystyle 5\) appear in the answer
 
logistic_guy said:
it say it correct, but it want me to express the answer in terms of the constant

for example if the differential equation is \(\displaystyle \frac{dy}{dx}\ln(\frac{dy}{dx}) = 5\), it want \(\displaystyle 5\) appear in the answer
Click to expand...
Sorry, but after almost 3 weeks I remember very little about that thread. I might try to refresh my memory later when I get time.
 
logistic_guy said:
it say it correct, but it want me to express the answer in terms of the constant

for example if the differential equation is \(\displaystyle \frac{dy}{dx}\ln(\frac{dy}{dx}) = 5\), it want \(\displaystyle 5\) appear in the answer
Click to expand...
It will help if you can show exactly what answer you gave, and exactly what "it" said in response, to make sure you aren't misinterpreting something, and that we know the details.

The mention of Lambert W in #18 may be relevant. Are you at all familiar with it? Have you looked it up?
 
logistic_guy said:
it say it correct, but it want me to express the answer in terms of the constant

for example if the differential equation is \(\displaystyle \frac{dy}{dx}\ln(\frac{dy}{dx}) = 5\), it want \(\displaystyle 5\) appear in the answer
Click to expand...
It is not difficult to write the solution in terms of the constant, but you have to understand the idea of the Lambert W function. It is similar to the idea of [imath]e^x[/imath] and [imath]\ln x[/imath] (a function and its inverse). To get the full picture of what I mean, watch this video and tell us what you did not understand:

 
You must log in or register to reply here.
Top