Differential equations: exact differential equation
The tutor explores how to detect and solve exact differential equations with a very simple example.
When a differential equation of the form
P(x,y) +Q(x,y)y’ = 0
results from the implicit differentiation of an original equation F(x,y)=c, the equation
P(x,y) +Q(x,y)y’ = 0
is said to be an exact differential equation.
The way to tell is that, for an exact differential equation,
Py(x,y) = Qx(x,y)
Example: Solve the differential equation
siny +1 +xy’cosy +2y’ = 0
First, we render it to P(x,y) +Q(x,y)y’ = 0:
siny + 1 +(xcosy +2)y’ = 0
P(x,y) = siny+1; Q(x,y) = xcosy + 2
Now we take the “other derivative” of each one:
Py(x,y) = cosy
Qx(x,y) = cosy
Py(x,y) = Qx(x,y): the equation is exact. Therefore,
siny +1 +xy’cosy +2y’ = 0
is the implicit derivative of some equation F(x,y) = c, which we need to find. Furthermore,
siny + 1 = Fx
We integrate siny + 1 with respect to x:
∫siny +1 = xsiny + x + g(y)
where g(y) is a function only of y that was lost in the original derivative by x.
g'(y) should be recognizable in Q(x,y): specifically, it’s 2⇒ g(y)=2y
Therefore, the solution F(x)=c is
xsiny +x +2y=c
Note that its implicit derivative is
siny + 1 + x(cosy)y’ +2y’ = 0
which matches the original differential equation posed.
Source:
Boyce, William and Richard DiPrima. Elementary Differential Equations and Boundary Value Problems. Toronto: John Wiley & Sons, 1986.
Jack of Oracle Tutoring by Jack and Diane, Campbell River, BC.
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