First order differentials equations

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Homework Help Overview

The discussion revolves around solving a homogeneous differential equation using the substitution z = y/x. Participants are exploring the transformation of the equation and the subsequent steps to express y in terms of x.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the transformation of the original equation and the implications of separating variables. There is a focus on the integration process and the correct identification of the variable of integration.

Discussion Status

Some participants have provided guidance on separating variables and questioning the integration steps. There is an ongoing exploration of the implications of the transformation and the integration process, with no explicit consensus reached yet.

Contextual Notes

There is mention of confusion regarding the variable of integration and the steps taken to rearrange the equation. Participants are also reflecting on potential errors in their methods and the implications of their transformations.

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Using z = y/x to transform the given homogeneous differential equation into a differential equation in z and x. By first solving the transformed equation, find the general solution of the original equation, giving y in terms of x.

[itex]z = \frac{y}{x} \rightarrow y = xz \rightarrow \frac{dy}{dx} = x\frac{dz}{dx} + z[/itex]
[itex]x\dfrac{dz}{dx} + z = \dfrac{x^3 + 4y^3}{3xy^2}[/itex]
[itex]x\dfrac{dz}{dx} + z = \dfrac{x^3 + 4z^3x^3}{3xz^2x^2}[/itex]
[itex]\dfrac{d}{dx}(xz) = \dfrac{1 + 4z^3}{3z^2}[/itex]
[itex]xz = \displaystyle\int (\dfrac{1 + 4z^3}{3z^2} )[/itex]
[itex]xz = \frac{-1}{3z} + \frac{2}{3}z^2 + c[/itex]
subbing z = y/x
[itex]x = \frac{-x^2}{3y^2} + \frac{2y}{3x} + \frac{cx}{y}[/itex]

How do I rearrange from y from here? I just can't seem to do it :\
 
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phospho said:
Using z = y/x to transform the given homogeneous differential equation into a differential equation in z and x. By first solving the transformed equation, find the general solution of the original equation, giving y in terms of x.

[itex]z = \frac{y}{x} \rightarrow y = xz \rightarrow \frac{dy}{dx} = x\frac{dz}{dx} + z[/itex]
[itex]x\dfrac{dz}{dx} + z = \dfrac{x^3 + 4y^3}{3xy^2}[/itex]
[itex]x\dfrac{dz}{dx} + z = \dfrac{x^3 + 4z^3x^3}{3xz^2x^2}[/itex]
In the next step, instead of writing the left side as d/dx(xz), subtract z from both sides. Then you will have ## x \frac{dz}{dx}## on one side and whatever you get on the right side. At this point you can separate the equation so that one side has all the stuff involving x and dx, and the other has all the stuff involving z and dz.
phospho said:
[itex]\dfrac{d}{dx}(xz) = \dfrac{1 + 4z^3}{3z^2}[/itex]
The step below is flaky. What is the variable of integration - dx or dz?
phospho said:
[itex]xz = \displaystyle\int (\dfrac{1 + 4z^3}{3z^2} )[/itex]
[itex]xz = \frac{-1}{3z} + \frac{2}{3}z^2 + c[/itex]
subbing z = y/x
[itex]x = \frac{-x^2}{3y^2} + \frac{2y}{3x} + \frac{cx}{y}[/itex]

How do I rearrange from y from here? I just can't seem to do it :\
 
Mark44 said:
In the next step, instead of writing the left side as d/dx(xz), subtract z from both sides. Then you will have ## x \frac{dz}{dx}## on one side and whatever you get on the right side. At this point you can separate the equation so that one side has all the stuff involving x and dx, and the other has all the stuff involving z and dz.
The step below is flaky. What is the variable of integration - dx or dz?

I seem to get the right answer by separating the variables as you mentioned.

The variable of integration is dz (wont let me edit now...), is there an error in my original method?
 
phospho said:
I seem to get the right answer by separating the variables as you mentioned.

The variable of integration is dz (wont let me edit now...), is there an error in my original method?
It doesn't seem right time.

Starting from this step:
$$ d/dx(xz) = \frac{1 + 4z^3}{3z^2}$$

What you're saying is that the derivative of xz with respect to x is <stuff on right side>, so xz must be the antiderivative of <stuff on right side> with respect to x. You can't just stick a dz in there out of nowhere.
 

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