Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
Solving a differential equation is not the same as an indefinite integral. – Thomas Andrews Apr 28 '13 at 22:45
With differential equations, you usually gain the constant from initial or boundary conditions. Do you have those? – icurays1 Apr 28 '13 at 22:45
What exactly are you trying to integrate in your differential equation example? – wj32 Apr 28 '13 at 22:46
The OP wants to find $\displaystyle \int \limits_{a}^{b}y'(x)dx$, for some $a,b$ that make sense. – Git Gud Apr 28 '13 at 22:47
up vote 2 down vote accepted

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.

share|cite|improve this answer
Oh thanks I see. And the cause of $y$ not being well defined in the first place is because $dy/dx$ is also not well defined. If $y=(x^2+1)e^C$, then $dy/dx=2(e^C)x$ – Ovi Apr 28 '13 at 23:08
@Ovi I don't like your choice of words, but you have right idea :) – Git Gud Apr 28 '13 at 23:14

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 $y(b)-y(a)$.

share|cite|improve this answer

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 $C$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.