WOLFRAM|DEMONSTRATIONS PROJECT

The Murder Mystery Method for Identifying and Solving Exact Differential Equations

​
ℳ
3

++2
2



3

+
2

2
+
2
3

3
The crime scene: ℳ d +  d = 0
​
3

+2
2

+​d + ​
2
3

3
+
2

2
+
3

​d = 0
Interrogate witness ℳ
Interrogate witness 
ℳ
3

+2
2

+

2
3

3
+
2

2
+
3

∫ℳ = ​
2
3


3
+
2


2
​ + ​
4

4
​ +

1
()
∫ = ​
2
3


3
+
2


2
​ + ​
3

3
​ +

2
()
Clues are consistent.
Implicit solution of the equation:
(
2
3


3
+
2


2
+ ​
4

4
+ ​
3

3
) = C
Finding out whether a first-order differential equation
Mdx+Ndy=0
is exact or not is solving a little "mystery": Is there a function
F(x,y)
such that its differential change
dF
is precisely
dF=Mdx+Ndy
? If the answer is "yes", then the equation implies that the change of the function
dF
is zero, and therefore the function is equal to a constant,
F(x,y)=C
. This last equality is actually an implicit solution of the differential equation
Mdx+Ndy=0
.
Tevian Dray and Corinne A. Manogue designed a fun and engaging way to explain how to solve these kind of "mysteries": A crime has been committed (the differential equation), and the student is the detective who must identify the murderer
F(x,y)
. The detective interrogates (integrates) the witnesses (
M
and
N
). If the clues given by them are consistent, then the murderer can be identified (therefore the implicit solution to the differential equation is obtained). If the clues are not consistent, the equation is not exact and has to be solved by another method.
Use the popup menus of this Demonstration to generate different differential equations and see this "Murder Mystery Method" applied to them. Pay attention to the colors and labels of the murderer's clothes and the mathematical expressions, as they show the relationship between the "murder" story and the mathematical procedure.
Elementary my dear Watson!