A differential equation is exact if it is of the form
#partial/{partialx}[F(x,y)]dx+partial/{partialy}[F(x,y)]dy=0#
If the differential equation is exact, then we would have
#partial/{partialx}F{x,y}=5x+4y# and #partial/{partialy}F{x,y}=4x-8y^3#
The order of partial derivatives on #F(x,y)# shouldn't matter so we must check whether the above conditions satisfy the equation below
#partial/{partialy}[partial/{partialx}F{x,y}]=partial/{partialx}[partial/{partialy}F{x,y}]#
Substitute,
#partial/{partialy}[5x+4y]quad?=?\quadpartial/{partialx}[4x-8y^3]#
#4 \quad ?=?\quad4#
Both sides are equal. Therefore there is a function, #F(x,y)#, that can meet the requirements that
#partial/{partialx}F{x,y}=5x+4y# and #partial/{partialy}F{x,y}=4x-8y^3#
and the differential equation is exact.
#partial/{partialx}F{x,y}=5x+4y#
#\impliesF(x,y)=5/2x^2+4xy+G(y)#
where #G(y)# is a function of #y# only. When integrating, we pick up an arbitrary constant. Since this is a partial derivative with respect to #x#, #y# is considered constant and the arbitrary constant can contain terms with #y#.
#partial/{partialy}F{x,y}=4x-8y^3#
#\implies F(x,y)=4xy-2y^4+Z(x)#
where #Z(x)# is a function that only depends on #x#
Comparing the two, we can set # G(y)=-2y^4+C# and #Z(x)=5/2x^2+C#
where C is a constant, and then the two expressions for #F(x,y)# would agree with each other. The solution to the exact differential equation is then.
#F(x,y)=5/2x^2+4xy-2y^4+C#
