I was solving a separable differential equation from my textbook and when I got the answer I verified that my answer was in fact a solution of the original D.E. but when I looked at the answer in the book looked a little bit different from mine.

This is the original D.E. :
xy' + y = y^2

y = c/(c-x)

This is the answer in the book
y = 1/(1-cx); also y=0 and y=1 are singular solutions.

Honestly I still don't know how to check if there are singular solutions, I didn't quite understand the explanation in the book, is it easy, ? And as for the general solution how does y = c/(c-x) become the same as y = 1/(1-cx)?
If I divide numerator and denominator by c then I get y = 1/(1-(x/c)), but is x/c just the same as cx? and if so is this something that applies only when dealing with differential equations?

For all singular solutions, $y'=0$. So, you are solving the quadratic equation $x(0)+y=y^2 \Rightarrow y(y-1) = 0 \Rightarrow y=0\text{ or }y=1$.

For $y = \dfrac{1}{1-\tfrac{x}{c}}$, let $C = \dfrac{1}{c}$. Then $y = \dfrac{1}{1-Cx}$. The two solutions differ only by the representation of the constant, so they are equivalent. It applies whenever you are dealing with arbitrary constants, so it arises in particular in differential equations, but may arise elsewhere (I cannot think of an example off the top of my head where else it might show up).

Originally Posted by catenary
I was solving a separable differential equation from my textbook and when I got the answer I verified that my answer was in fact a solution of the original D.E. but when I looked at the answer in the book looked a little bit different from mine.

This is the original D.E. :
xy' + y = y^2

y = c/(c-x)

This is the answer in the book
y = 1/(1-cx); also y=0 and y=1 are singular solutions.

Honestly I still don't know how to check if there are singular solutions, I didn't quite understand the explanation in the book, is it easy, ? And as for the general solution how does y = c/(c-x) become the same as y = 1/(1-cx)?
If I divide numerator and denominator by c then I get y = 1/(1-(x/c)), but is x/c just the same as cx? and if so is this something that applies only when dealing with differential equations?
The equation is separable, so you have

\displaystyle \begin{align*} x\,\frac{\mathrm{d}y}{\mathrm{d}x} + y &= y^2 \\ x\,\frac{\mathrm{d}y}{\mathrm{d}x} &= y^2 - y \\ \frac{1}{y^2 - y} &= \frac{1}{x} \end{align*}

which is now able to be integrated.

Notice that you can only divide by these terms as long as they are not 0, so that means that your solutions will be correct provided \displaystyle \begin{align*} x \neq 0, y \neq \left\{ 0, 1 \right\} \end{align*}

So that means you need to check if these also give solutions to the equation. Substitute them in and see if you get a true equation...

Originally Posted by Prove It
Substitute them in and see if you get a true equation...
Substitute them in what? Define 'true equation'? Sorry if it may seem like trivial questions but I really am struggling with the jump from calculus to differential equations

Your starting equation \displaystyle \begin{align*} x\,\frac{\mathrm{d}y}{\mathrm{d}x} + y = y^2 \end{align*}...

Originally Posted by Prove It
Your starting equation \displaystyle \begin{align*} x\,\frac{\mathrm{d}y}{\mathrm{d}x} + y = y^2 \end{align*}...
So when dealing with differential equations, is there a rule that once you have separated the equation, if there are numbers that are outside of the domain and range of the separated form (for the case above x=0 and y={0,1}), you have to substitute those numbers in the original equation before it was separated because those might give you singular solutions?
a.) x=0
$0 + y = y^{2}$
$y((y - 1) = 0$
So that is what you mean by a true equation?
Solving for y we get y = 0 and y = 1
b.) y=0
I get $xy' = 0$ What am I supposed to do with this? Or is y' supposed to be equal to 0 because I made y a constant so it turns out to be 0 = 0? So what would this mean? I guess this is not a true equation so this is disregarded?
c.) y=1
is it $xy' + 1 = 1$
$xy' = 0$
or is it
y=1 therefore y'=0 therefore
$0 + 1 = 1$
$1 = 1$
And this also is not a true equation?

Originally Posted by catenary
So when dealing with differential equations, is there a rule that once you have separated the equation, if there are numbers that are outside of the domain and range of the separated form (for the case above x=0 and y={0,1}), you have to substitute those numbers in the original equation before it was separated because those might give you singular solutions?
Yes that is exactly right.

a.) x=0
You don't need to test x = 0 because the solutions to a DE are the y functions of x which satisfy the equation.

b.) y=0
I get $xy' = 0$ What am I supposed to do with this? Or is y' supposed to be equal to 0 because I made y a constant so it turns out to be 0 = 0? So what would this mean? I guess this is not a true equation so this is disregarded?
Yes y = 0 so y' = 0, thus the LHS is xy' + y = 0 and the RHS is y^2 = 0, so LHS = RHS, this equation IS true. Thus y = 0 is a solution.

Try the same thing for y = 1.

Originally Posted by Prove It
Try the same thing for y = 1.
I did try the same thing for y = 1 already in my post:
c.) y=1
is it xy' + 1 = 1
xy' = 0
or is it
y=1 therefore y'=0 therefore
0 + 1 = 1
1 = 1
And this also is not a true equation?
So I guess since LHS = 1 and RHS = 1 then it qualifies as a true equation.. Now I understand thank you proveit