Dy/dx = (y^2) /x , form differential equation

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Homework Help Overview

The discussion revolves around the differential equation dy/dx = (y^2)/x. Participants are examining their solutions and comparing them to a provided answer, exploring the implications of their findings.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss their solutions and question the validity of the given answer. There is an exploration of the correct form of the solution, particularly regarding the placement of constants and the use of absolute values in logarithms.

Discussion Status

The conversation is ongoing, with some participants affirming the correctness of their solutions while others suggest potential errors in the provided answer. There is a focus on ensuring clarity in the problem statement and the correct application of mathematical principles.

Contextual Notes

Participants note the importance of including the original differential equation in the body of the posts rather than the title. There is also mention of arbitrary constants and their roles in the solutions presented.

goldfish9776
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Homework Statement


my ans is lnx = (-1/y) + c
(-1/y) = lnx -c
y = -1/ (lnx -c ) , but the answer given is (-1/ln x )+ C , how to get the answer given ?

Homework Equations

The Attempt at a Solution

 
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Your solution seems fine. I believe the given answer may be wrong.
 
You can easily verify that your answer is correct and the given answer is wrong by plugging them back into the original DE. By the way, you should post the original DE in the body, not the title of your post.

The only thing I would change is use ##\ln |x|## in the antiderivative.
 
goldfish9776 said:

Homework Statement


my ans is lnx = (-1/y) + c
(-1/y) = lnx -c
y = -1/ (lnx -c ) , but the answer given is (-1/ln x )+ C , how to get the answer given ?

Homework Equations

The Attempt at a Solution

When you post a question, please put the problem statement into the first section, not the thread title.

Your answer looks fine to me, except that it should be ln|x|, but you should carry it a step further and solve for y. The given answer should have the constant in the denominator, not just added to the fraction as you show it.
 
And more often than not for a result like that you would be writing, since c is an arbitrary constant

y = -1/(ln x + ln K) where K is another arbitrary constant

= -1/(ln Kx) and thence maybe

x = K' e-y
 

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