Simple differential equation - Finding the solution

dumbdumNotSmart
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Homework Statement


We are given the following differential equation:

$$ y+yy'-xy'=0 $$

Let's find the solution.

Homework Equations

The Attempt at a Solution


So in my course usually we have to do some sort of substitution by the lines of x/y=z or y/x=z. This equation has proven difficult. I have taken a look at wolfram alpha and the y(x)=xv(x) substitution is not one I am familiar with.
 
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##y + (y - x ) y' = 0##
Can be written as:
##y' = -\frac{y}{y-x} = -\frac{1}{1 - x/y} ##
So, if you were to use ## z = x/y,## you would get ##z' = \frac{yx' - xy'}{y^2}##.
Then substituting into the original equation would give yy' + y^2 z' = 0.
Assume y is not zero, and you get:
##\frac{y'}{y} = -z' ##
Integrate both sides.
 
Could you give me a more chew down process of the substitution? What goes where to get that lovely equation down there at the bottom of your answer?
 
dumbdumNotSmart said:
Could you give me a more chew down process of the substitution? What goes where to get that lovely equation down there at the bottom of your answer?
I think he did show how he got it.

Work through a bit of algebra.
 
dumbdumNotSmart said:
Could you give me a more chew down process of the substitution? What goes where to get that lovely equation down there at the bottom of your answer?
Here is a bit more information to guide your algebra.
##z' = \frac{yx' - xy'}{y^2} = \frac{1}{y^2}( y - xy') ##. So ##y^2 z' = y - xy'## which matches 2 of the 3 terms in your original equation.
So, substituting into the original equation would give ##yy' + y^2 z' = 0##.
Assume y is not zero, and divide the whole thing by ##y^2##, you get:
##\frac{y'}{y} + z'=0 ## which is the same as ##\frac{y'}{y} = -z' ##.
 
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