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**Case 2**: *r*_{1} ≠ *r*_{2}, *k*_{1}=0

Here the rate of the inflow and the outflow differs. The volume will thus be

This will give us

This is also a separable equation. We get

Integrating both sides will give us

(Yet agin we don’t need to care about the absolute sign since we cannot have negative volumes.)

This gives us

or

This might look fairly messy, but as we can see, given reasonable numbers, it will turn out rather OK.

**An example**

Say we have *V*_{0}=2000 l, *r*_{1} =15 l/s, *r*_{2} = 20 l/s, *k*_{1}=0 kg/l and *y*_{0} = 50 kg. This would give us the solution

Finally, we need to find C, but to make that somewhat easier we can first move out a factor of 2000^{4} out of the bracket.

We can then let the constant *C*_{2} absorb the 2000^{4} to get us

Let us compare this to example solution for case 1. The red curve is our case 2.

We can here see that the solutions are quite similar, but that the amount of salt goes to 0 more quickly. After 2000/5=400 second the volume is 0, and we could thus not have any salt left.

**Back to the general solution**

We can go back to our general solution to get

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