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I have been working on solving differential equations and this is really cracking me up. I obtained the following equation:

dz/dr = r

and I wish to obtain z in terms of r, given that we know that z = 0 when r = 0; is this:

$dz = r dr$

$\int_0^z 1\,dz = \int_0^r r \,dr$

$==> z = r^2/2$

the right way of doing this(are the limits of the integration correct)?

thank you

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That's right. And it's the simplest way to do these initial value type questions. – coffeemath Nov 11 '12 at 2:29
up vote 1 down vote accepted

As @coffeemath said, this is definitely a fine way to do the question like this, but I agree: the fact that the limits of integration are different on each side is a bit unsettling (we would like to solve problems by only doing the same thing to both sides of an equation).

It may be more illustrative to do the problem in as slightly different way to see what's going on. In this context, $z$ is shorthand for $f(r)$, where $f$ is some function. $\frac{\mathrm{d}z}{\mathrm{d}r}=r$ really means $f'(r)=r$. Then we can apply $\int_0^R\cdot\mathrm{d}r$ to both sides of that equation: $\int_0^Rf'(r)\mathrm{d}r=\int_0^R r\mathrm{d}r$. On the left hand side we have $f(R)-f(0)$ by the fundamental theorem of calculus, and on the right hand side we have $R^2/2$. Finally, since we know $f(0)=0$ (since $z=0$ when $r=0$), this just says $f(R)=R^2/2$. This should work for any $R$, so we can write $f(r)=r^2/2$ or just $z=r^2/2$.

This is essentially the same calculation you did, but be careful, your original calculation contained two errors in notation that cancelled out. If you write $\int_0^z 1\mathrm{d}z$, it should evaluate to something like "the value of $z$ at $r=$'$z$' minus the value of $z$ at $r=0$ (which happens to be $0$)", which isn't quite what you want. By writing $z$ on the next line instead of something more like $z\mid_{r=z}$ you made a second mistake (in notation) that fixed the issue.

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