Integrating factors or separating the variables

In summary, the equation \frac{dy}{dx} - \frac{y}{4x} = 0 can be solved by either integrating factors or separating the variables. After integrating both sides, the equation can be simplified to y = x^1/4, with the initial condition of y(2)=3. The constant should be added after integration and then multiplied after exponentiation to get the final solution.
  • #1
issisoccer10
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[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
 
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  • #2
It looks pretty separable to me. dy/dx=y/(4x), dy/y=dx/(4x). Now just integrate both sides.
 
  • #3
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..
 
  • #4
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.
 
  • #5
alright thank you.. so the constant has to be added right after the integration occurs.. I forgot about that. thanks a lot
 

Related to Integrating factors or separating the variables

1. What are integrating factors?

Integrating factors are mathematical tools used to solve certain types of differential equations, specifically those that are not exact or separable. They allow us to transform a differential equation into one that is easier to solve.

2. How do integrating factors work?

Integrating factors work by multiplying both sides of a differential equation by a specific function that helps to make the equation exact. This function is known as the integrating factor, and it is chosen based on the form of the differential equation.

3. When should integrating factors be used?

Integrating factors should be used when solving certain types of differential equations, such as those that are not exact or separable. They are also useful for solving equations with variable coefficients.

4. What is the purpose of using integrating factors?

The purpose of using integrating factors is to simplify the process of solving differential equations. By transforming the equation into an exact form, we can use basic integration techniques to find the solution.

5. Are there any limitations to using integrating factors?

There are some limitations to using integrating factors, as they can only be applied to certain types of differential equations. Additionally, finding the correct integrating factor can be a difficult and time-consuming process, especially for more complex equations.

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