# Simple differential equation - Finding the solution

1. Dec 1, 2015

### dumbdumNotSmart

1. The problem statement, all variables and given/known data
We are given the following differential equation:

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

Let's find the solution.
2. Relevant equations

3. 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.

2. Dec 1, 2015

### RUber

$y + (y - x ) y' = 0$
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.

3. Dec 1, 2015

### dumbdumNotSmart

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?

4. Dec 1, 2015

### SammyS

Staff Emeritus
I think he did show how he got it.

Work through a bit of algebra.

5. Dec 2, 2015

### RUber

$z' = \frac{yx' - xy'}{y^2} = \frac{1}{y^2}( y - xy')$. 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 $\frac{y'}{y} = -z'$.