# 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,

P_{y}(x,y) = Q_{x}(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:

P_{y}(x,y) = cosy

Q_{x}(x,y) = cosy

P_{y}(x,y) = Q_{x}(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 = F_{x}

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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