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I am trying to integrate a simple function:

$\frac{dA}{dt} = j - kA(t)$

where j and k are constants, and $A(0)=A_0$ is also a constant. I got to

$A(t) = jt + C - k\int A(t)dt$

but I dont know how to proceed now. Could you help me out please? I know the answer is

$A(t) = e^{-k*t} (A_0+ \frac{j}{k}e^{k*t} -1)$

How can I get there? thanks so much

2 Answers 2

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If $k=0$, this is is easy. Suppose $k\neq 0$.

Let's first rewrite it as $\frac{dA}{dt}+kA=j.$ Multiplying both sides by the positive function $e^{kt}$ gives us $e^{kt}\frac{dA}{dt}+ke^{kt}A=je^{kt}.$ Using the product rule on the left side, we have $\frac{d}{dt}\left[e^{kt}A\right]=je^{kt}.$ Integrating with respect to $t$, we then have $e^{kt}A=j\int e^{kt}\,dt=\frac{j}{k}e^{kt}+C,$ and so $A(t)=\frac{j}{k}+Ce^{-kt}.$ Using our initial value to find $C$, we get $A(t)=\frac{j}{k}+\left(A_0-\frac{j}{k}\right)e^{-kt}.$

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    LaTeX is the preferred method to use. If you're ever curious about a particular formatted bit, right-click on it, then choose Show Math As-->TeX Commands.2012-11-13
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Do you know that the solution to $\frac {dA}{dt}=-kA$ is $A(0)\exp^{-kt}$? You can see that (not a proof) by differentiating and plugging in. If so, define $B(t)=A(t)-\frac jk$. Then the answer you give just pulls out the factor $e^{-kt}$

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    thanks Ross! I'll check it out!2012-11-12