When solving a homogenous DE, I am suppose to find a variable to substitute into the DE such as
$$ let , u=\frac{y}{x}$$
Im wondering if can have multiple answers to one DE depending on what I choose as the substitute?
For example:
$$ \frac{dy}{dx} =\frac{y-x}{y+x} $$ $$ let , u=y+x$$ $$ then , du=\frac{dy}{dx} +1$$
I ended up getting $$ -y-x= x+C$$, however the solution in the back of the book has ln's. Which is why Im wondering if there are more than one solution to a problem like this.
First Case
$$u = \dfrac{y}{x} \implies y = u x \implies y' = u + u' x = \dfrac{\dfrac{y}{x} - 1}{\dfrac{y}{x}+1} = \dfrac{u-1}{u+1}$$
Solving the Separable Equation, we get
$$\dfrac{1}{2} \ln(u^2 + 1) + \arctan(u) = -\ln x + c$$
Substitute $u(x) = \dfrac{y}{x}$.
Second Case
$$u = y + x \implies y = u - x \implies y' = u' - 1 = \dfrac{y-x}{y+x} = \dfrac{u - 2 x}{u}$$
To solve this, we use the substitution, $u = x y$ and get the same result after.
Do you see how the first method was much better as it reduced the DEQ to a separable one without having to make a second substitution, which takes you back to the first case anyway?
Regardless, you get the same result.