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Consider the differential equation with $\frac{dx}{dt}=1-x$ and $x=0$ when $t=0$.

The answer uses the result that $\int \frac{dx}{1-x}=\ln (1-x)$, hence getting the solution $x=1-e^{-t}$.

However I use $\int \frac{dx}{1-x}=\ln \left|1-x\right|$ instead, which gets me $x=1-e^{-t} \text{ or }1+e^{-t}$.

Am I right, or is it standard to not use the absolute signs?

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    @ThomasE. It seems to me that you can solve your equation only in the connected region where the initial condition belongs. In this case $x(0)=0$, so that any maximal solution will stay below the line $x=1$. Am I wrong?2012-04-23

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$ \frac{dx}{dt} = 1-x $ $ \frac{dx}{1-x} = dt $ $ -\ln|1-x| = t+\text{constant} $ $ |1-x| = e^{-t}\cdot\text{positive constant} $ (Since $e$ to a real power is always positive.) $ 1-x=e^{-t}\cdot\text{constant} $ $ x = 1 - e^{-t}\cdot\text{constant} $ Since we want $x=0$ when $t=0$, the "constant" is $1$.

If you had wanted $x$ to be $2$ when $t=0$, then you'd have been in trouble if you'd neglected the absolute value.

All of this works if $x\ne 1$. But you can check that $x=1$ for all values of $t$ is also a solution.

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    Good. But the absolute value is indeed useless, since $x(0)=0$ and x(t) < 1 by the local uniqueness theorem. Anyway, your solution is standard and works for every initial condition.2012-04-23