# The answer to the following: dy/dx = y + 2 leads to: ln|y+2| = x + C and the solution is y = -2 + Ce^x How does e ^ (x+C) equate to Ce ^x

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It doesn't. In general `e^(x+C)!=Ce^x.` What's really happening is that since `e^(x+C)=e^xe^C,` and since `e^C` is also a constant whenever `C` is, we can write `e^xe^C=C_1e^x,` where `C_1=e^C.` But there's no particular reason for our choice of labels (we could have written the first constant as `C_1` and then let `C=e^(C_1)` ), so it's easier to drop the subscript and just write `Ce^x` instead -- it's just that this `C` isn't the same as the `C` in the previous step.

Actually, writing `Ce^x` instead of just `e^(x+C)` gives more solutions, since `y=-2-e^x` is a solution that can't be written in the form `-2+e^(x+C).` So if anyone misses the solutions `y+2=-e^(x+C)` while dropping the absolute value bars, and the trivial solution `y-=-2`, they'll recover them when making the switch at the end.

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