[spec] concentration integration and separation of variables

diff --git a/spec/calculus.md b/spec/calculus.md
index 43cf752fc9da76c5fece553dbecaeed0be5d433f..97bc926a7cb6b59e66ec2eba6c634ac742cb6028 100644 (file)
--- a/spec/calculus.md
+++ b/spec/calculus.md
@@ -398,3 +398,13 @@ Start with $y=\dots$, and differentiate. Substitute into original equation.
 ### Function of the dependent variable
 
 If ${dy \over dx}=g(y)$, then ${dx \over dy} = 1 \div {dy \over dx} = {1 \over g(y)}$. Integrate both sides to solve equation. Only add $c$ on one side. Express $e^c$ as $A$.
 ### Function of the dependent variable
 
 If ${dy \over dx}=g(y)$, then ${dx \over dy} = 1 \div {dy \over dx} = {1 \over g(y)}$. Integrate both sides to solve equation. Only add $c$ on one side. Express $e^c$ as $A$.

+
+### Mixing problems
+
+$$\left({dm \over dt}\right)_\Sigma = \left({dm \over dt}\right)_{\text{in}} - \left({dm \over dt}\)_{\text{out}}$$
+
+### Separation of variables
+
+If ${dy \over dx}=f(x)g(y)$, then:
+
+$$\int f(x) \> dx = \int {1 \over g(y)} \> dy$$