Homogeneous differential equation - serious help

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Homework Help Overview

The discussion revolves around solving a homogeneous differential equation using the substitution \( v = \frac{y}{x} \). The original poster presents the equation \( \frac{dy}{dx} = \frac{3y^2 - x^2}{2xy} \) and attempts to manipulate it into a more workable form.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the manipulation of the differential equation, with some questioning the correctness of the original poster's steps. There is an emphasis on expressing both sides of the equation in terms of the substitution variable \( v \) and its derivative.

Discussion Status

Some participants have provided guidance on how to proceed with the substitution and have pointed out errors in the original poster's calculations. The conversation indicates progress in understanding the problem, though there is no explicit consensus on the final approach.

Contextual Notes

There was a correction made to the original equation, which was initially misstated. The discussion also highlights the importance of retaining the original problem for educational purposes.

masterchiefo
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Homework Statement


I need to resolve this with v = y/x

dy/dx= (3y2-x2)/(2xy)

Homework Equations

The Attempt at a Solution


dy/dx= (3y2-x2)/(2xy)

dy/dx= 3y2/2xy -x2/2xy

dy/dx = 3y/2x -x/2y

dy/dx = 3y/2x - 1/2y/x

dy/dx = 3/2 *v - 1/2*v

F(v) = 3/2 *v - 1/2*v

is that good so far ?
 
Last edited:
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The First term in your 2nd line is wrong, and the error propagates through from there.
 
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andrewkirk said:
The First term in your 2nd line is wrong, and the error propagates through from there.
sorry I made a mistake in the original equation and now its fixed.
it was 3y3 but its actually 3y2
 
You've expressed the right-hand side in terms of v, so that's progress. Now you need to do similar work on the left hand side.

Using ##v=\frac{y}{x}##, express ##\frac{dv}{dx}## in terms of y, y' and x, then see if you can use that to express y' in terms of v, v' and x. Equate that to the RHS and then with any luck you'll be able to use separation of variables to get an expression to be integrated over x on one side and the same for v on the other.
 
andrewkirk said:
You've expressed the right-hand side in terms of v, so that's progress. Now you need to do similar work on the left hand side.

Using ##v=\frac{y}{x}##, express ##\frac{dv}{dx}## in terms of y, y' and x, then see if you can use that to express y' in terms of v, v' and x. Equate that to the RHS and then with any luck you'll be able to use separation of variables to get an expression to be integrated over x on one side and the same for v on the other.
what do you mean by left hand side ?
 
LHS of the diff equation in section 1 of your OP, which is: ##\frac{dy}{dx}##.
 
andrewkirk said:
LHS of the diff equation in section 1 of your OP, which is: ##\frac{dy}{dx}##.
hey man, thanks for your time and help, I was able to get it resolved and it matched the answer in my book.
you are awesome.
 
Please do NOT erase the original problem after it has been solved. Other people can learn from this.
 

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