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$\frac{dP}{dt} = P - a$ assume that $a$ is a contant.

My solution.

$dP = \left(P-a\right)dt$

$\int \frac{\mathrm{d}P}{P-a} = \int \mathrm{d}t$

$\ln \vert P-a \vert = t + C$

$P-a = Ce^{t}$

$P = Ce^{t}+a$

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It is correct. To check, differentiate and check that the differential equation is satisfied. Then check that the initial condition is satisfied (in this case you have no initial condition). – copper.hat Feb 22 '13 at 7:09
If you just want to check your answer you can also try Wolfram|Alpha: – in_wolframAlpha_we_trust Feb 22 '13 at 7:24
up vote 1 down vote accepted

Your answer is pefectly all right.If you write some more steps it will be clear to you that it is all right.

Steps like ,

Let $x=P-a$ then we have $dx=dP$ so the integral becomes $\int\frac{dx}{x}=\int dt$

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