The discussion revolves around solving a separable differential equation given by (y-1)dx + x(x+1)dy = 0. Participants are exploring the integration process and the resulting expressions.
There is ongoing exploration of the relationship between different forms of the solution, with some participants suggesting that variations in the constant do not indicate an error in the approach. Clarifications about the treatment of differentials are also being discussed.
Some participants reference solutions from textbooks and computational tools, indicating a comparison of results. There is mention of potential misunderstandings regarding the manipulation of the equation during the integration process.
Arman777 said:Homework Statement
##(y-1)dx+x(x+1)dy=0##
Homework Equations
The Attempt at a Solution
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I multiplied both equation with, ##\frac {1} {(y-1)x(x+1)}## so I get
##\frac {dx} {x(x+1)}+\frac {dy} {y-1}=0##
taking integral for both sides
then I get
##ln(x)-ln(x+1)+ln(y-1)=ln(c)##
so
##ln(\frac {x(y-1)} {x+1})=ln(c)##
then I get ##xy-x=c(x+1)##
Anyone who can notice where I am doing wrong ?
Im my book it says ##xy+1=c(x+1)## also in wolframDick said:What makes you think something is wrong?
Arman777 said:Im my book it says ##xy+1=c(x+1)## also in wolfram
fresh_42 said:The multiplication by ##\frac {1} {(y-1)x(x+1)}## affects the differentials. You cannot treat it as a constant.
I made a mistake, sorry.Dick said:Hmm? In what way does it affect the differentials?
fresh_42 said:I made a mistake, sorry.
So, I did right then Since ##c-1## and ##c## are same.Dick said:It's the same solution with a different choice of ##c##. Substitute ##c-1## for ##c## in your solution and you'll get the book solution.
Arman777 said:So, I did right then Since ##c-1## and ##c## is same.