From befe17c800bf6752581f9e2eb53e8c05d1275ef3 Mon Sep 17 00:00:00 2001
From: Andrew Lorimer <andrew@lorimer.id.au>
Date: Tue, 21 May 2019 12:22:28 +1000
Subject: [PATCH] [spec] concentration integration and separation of variables

---
 spec/calculus.md | 10 ++++++++++
 1 file changed, 10 insertions(+)

diff --git a/spec/calculus.md b/spec/calculus.md
index 43cf752..97bc926 100644
--- 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 $A$.
+
+### 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$$
-- 
2.43.2