Question

There exists a function u = u(x,y) such that du = Mdx + Ndy where M and N are functions of x and y.

Then which of the following option is always correct for the differential equation dx + Ndy = 0 ?​

Answer 1

Answer:

Note that M, N ∈ C

1

(R) and My = 3 = Nx. Thus, there

exists f(x, y) such that fx = 4x + 3y and fy = 3x + 3y

2

.

fx = 4x + 3y ⇒ f(x, y) = 2x

2 + 3xy + ϕ(y). Now,

3x + 3y

2 = fy(x, y) = 3x + ϕ

′

(y).

⇒ ϕ

′

(y) = 3y

2 ⇒ ϕ(y) = y

3

.

Thus, f(x, y) = 2x

2 + 3xy + y

3

and the general solution is

given by

2x

2 + 3xy + y

3 = C

Answer 2

Answer:

Definition: Let F be a function of two real variables such that

F has continuous first partial derivatives in a domain D. The

total differential dF of the function F is defined by the formula

dF(x, y) = Fx(x, y)dx + Fy(x, y)dy

for all (x, y) ∈ D.

Definition: The expression M(x, y)dx + N(x, y)dy is called an

exact differential in a domain D if there exists a function F

such that

Fx(x, y) = M(x, y) and Fy(x, y) = N(x, y)

for all (x, y) ∈ D.

Definition: If M(x, y)dx + N(x, y)dy is an exact differential,

then the differential equation

M(x, y)dx + N(x, y)dy = 0

is called an exact differential equation.

Step-by-step explanation:

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