Is this definite integral impossible? From my understanding when you integrate $f(x)$ you get $F(x)+C$, and when finding a definite integral the $C's$ cancels out due to subtraction. However, I came across an example where the $C$ doesn't cancel out: so I started with the following differential equation: $$(1+x^2) \frac{dy}{dx}=2xy$$ and suppose I wanted to find the area under $dy/dx$ between $a$ and $b$. All you simply have to do is find $y$, evaluate at $b$ and $a$, and subtract. The solution to this equation is $y=(x^2+1)e^C$.
Now if you evaluate and subtract, you get $(b^2+1)e^C-(a^2+1)e^C$. Is this integral impossible unless I have more information which allows me to determine $C$? Thanks!
 A: You already got that there exists $C\in \Bbb R$ such that for all $x\in I$, (where $I$ in some interval), $y(x)=(x^2+1)e^C$.
Allow me to change the variable for the sake of trying to enlighten you: there exists $K\in \Bbb R$ such that for all $x\in I$, $y(x)=(x^2+1)e^K$. (Forget about $C$).
It is true that $y$ is a fixed solution to the differential equation, period.
You wish to find $\displaystyle \int \limits_{a}^b y'(x) \,dx$, where $a,b\in I$.
You know $y$ is an antiderivative for $y'$. Therefore the set of antiderivatives for $y'$ (in the interval $I$) is $\left\{Y\in \Bbb R^I: (\exists C\in \Bbb R)(\forall x\in I)\left(Y(x)=y(x)+C\right)\right\}$.
Take any antiderivative $Y$ of $y$. There exists $C\in \Bbb R$ such that $Y(x)=y(x)+C=(x^2+1)e^K+C$.
And you know that $$\begin{align} \displaystyle \int \limits_{a}^b y'(x) \,dx&=Y(b)-Y(a)\\&=(b^2+1)e^K+C-(a^2+1)e^K-C \\
&=(b^2-a^2)e^K\end{align}$$
This is well defined, there's nothing wrong with it because $y$ was a fixed solution to the differential equation, which means $(b^2-a^2)e^K$ is a real number.

Javier on his answer says $y$ isn't completly specified. I say exactly the opposite, but we're not contradicting each other, we're using specified with different meanings.
This comes down to the same similar problem. Let $X=\{a,b,c\}$. Let $x\in X$.
I say $x$ is specified (or fixed) while Javier says it isn't speficied because we don't know what $x$ is, but both of us will work with $x$ in the same way.
A: Your constant doesn't come from integrating $y'$. Rather, it comes from the fact that since you haven't specified an initial condition, $y$ isn't completely specified from the differential equation. Take a look: you say that solution is $(x^2+1)e^C$. You haven't integrated $y'$ yet, but you still don't know $y$ until you pick a value for $C$. And when that happens, then you can evaluate $y(b)-y(a)$.
A: The purpose of solving an ordinary differential equation
(using the standard variables $x$ and $y$)
is to find the the value of $y$ for a specified $x$
given the differential equation
and the initial values.
Subtracting the values makes no sense in this case.
In your case, you have a family of solutions,
one for each value of $C$.
