Question #8dd37

1 Answer
Jun 9, 2015

The differential equation is exact. The solution is:
F(x,y)=5/2x^2+4xy-2y^4+C

Explanation:

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