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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!

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Solving a differential equation is not the same as an indefinite integral. –  Thomas Andrews Apr 28 '13 at 22:45
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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

3 Answers 3

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.

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

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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 $C$.

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