Question

Q.1 State and prove the necessary and sufficient condition for the equation
Mdx+Ndy=0 to be exact.


Q.2 Solve the equation
(P+y+x)(xp+y+x) = 0


0.4 Difine a partial differential equation and form a partial differential equation by eliminating the arbitrary functions f
and F from y = f(x-at) + F(x +at).​

Answer 1

State and prove the necessary and sufficient condition for the equation

Mdx+Ndy=0 to be exact.

Step-by-step explanation:

The necessary and sufficient condition for equation  is:

Mdx + Ndy = 0 (a)

to be exact is that

∂M /∂y = ∂N /∂x . (b)

Proof. Necessity: Suppose that (a) is exact.

∴, Mdx + Ndy can be obtained  directly by differentiating some function

f = f(x, y). Thus,

d[f(x, y)] = Mdx + Ndy ⇒ ∂f /∂x (dx )+ ∂f /∂y (dy) = Mdx + Ndy.

∴, ∂f /∂x = M and ∂f /∂y = N;

∂ ² f /∂y∂x = ∂M /∂y and ∂ ²f /∂x∂y = ∂N /∂x . (c)

∵, we assume the function is many times continuously differentiable,

, ∂ ²f /∂y∂x = ∂ ²f /∂x∂y .

∴ (c) gives ∂M /∂y = ∂N /∂x .

For Sufficiency. Let P = ∫ Mdx. Then ∂P /∂x = M.

∴ ∂ ²P ∂y∂x = ∂M /∂y .

The above equation together with  (b) gives:

∂N /∂x = ∂M /∂y = ∂ ²P /∂y∂x = ∂ ²P /∂x∂y = ∂ /∂x ( ∂P /∂y )

We get this, on integrating with respect to x,

gives,

N = ∂P /∂y + φ(y),

where φ is a function of y only. We will get:

Mdx + Ndy = ∂P ∂x( dx) + [ ∂P ∂y + φ(y) ] dy

= ∂P ∂x (dx) + ∂P /∂y (dy) + φ(y)dy

= dP + d(F(y)) (where d(F(y)) = φ(y)dy)

= d[P + F(y)],

The above proves that Mdx + Ndy = 0, is an exact differential equation