Integrating factors or separating the variables

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 3K views
issisoccer10
Messages
35
Reaction score
0
[SOLVED] Integrating factors or separating the variables

Homework Statement


The following equation can be solved by intergrating factors or by separating the variables.

[tex]\frac{dy}{dx}[/tex] - [tex]\frac{y}{4x}[/tex] = 0

with the initial condition of y(2)=3


Homework Equations





The Attempt at a Solution


This problem is the final part of a question in which I am supposed to find the trajectory of a particle if it moves continuously in the directoin of maximum temperature increase. Setting the trajectory so that it is in the direction of the gradient allows me to figure out all the way to the point above. However, I cannot simplify the equation down further into a function of y in terms of x... any help would be greatly appreciated... thanks
 
Physics news on Phys.org
integrating both sides gives ln(y) = ln(x)/4 = ln(x^1/4)... so y = x^1/4.. as for the inital condition, my inclination would be to plug in (2,3) into the equation... so 3 = 2^1/4 + C.. However, based on what I think it should be, instead of adding C I should be multiplying by C. But I don't know why..
 
You get ln(y)=ln(x^(1/4))+C. That's fine. But to get rid of the logs you exponentiate exp(ln(x^(1/4))+C)=exp(ln(x^(1/4))*exp(C)=x^(1/4)*exp(C). The additive constant becomes multiplicative after you exponentiate.
 
alright thank you.. so the constant has to be added right after the integration occurs.. I forgot about that. thanks a lot