Differential Equation Theorems And Explanations
As I have discussed previously, there are many methods and related theorems to solve a differential equation.
So now I will discuss all theorems and applications related to it.It may be longer but beneficial for solving problems of differential equations.Theorem_1: The differential equation
M(x,y) dx + N(x,y)dy= 0 is exact iff
∂M/∂y= ∂N/∂x
Proof:
If the given equation is exact , we have d(u(x,y))= M(x,y) dx + N(x,y) dy. ... (1)d(u(x,y)) = ∂u/∂x dx + ∂u/∂y dy .......(2)
Consequently by comparison of (1) and(2)
∂u/ ∂x= M(x,y) , ∂u/ ∂y = N(x,y) .........(3)
Moreover,
∂M/∂y=∂²u/ ∂y∂x and ∂N/∂x=∂²u/ ∂x∂y
and because of ∂²u/∂y∂x =∂²u/∂x∂y, we get
∂M/∂y=∂N/∂x ..........(4)
Convesly ,
let ∂M/∂y= ∂N/∂x and we shall prove that
M(x,y) dx + N(x,y) dy is an exact differential equation.
Let V (x,y) = ∫ M dx, Here we have integrated M( x,y) with respect to x treating y as a constant.
Then
∂V/ ∂x = M; ∂²V/∂x∂y= ∂M/∂y= ∂N/∂x. ...(5)
Therefore ∂N/ ∂x= ∂/∂x(∂V/∂y) .....(6)
Hence integrating w.r.t x, we obtain
N = ∂V/∂y + φ'(y), ......(7)
where φ'(y) is some function of y.
Thus from (5) and (7)
M(x,y) dx + N( x,y) dy
= ∂V/∂x dx+ ∂V/∂y dy +φ'(y) dy
= d[V+ φ'(y)], an exact differential.
( proved )
Solved Example:
Solve the differential equation (x+y+1) dx + (x - y²+ 3) dy =0if it is exact.
Solution:
Here M = x+ y + 1 and N = x - y² + 3∴ ∂M/∂y = 1 and ∂N/∂x = 1
since ∂M/∂y = ∂N/∂x, the given equation is exact.
Then its solution is given by
∫ M dx + ∫ N dy = c where c is a constant
| |y as const terms don' t
contain x
=> ∫( x+y+1 ) dx +∫ (-y² + 3) dy = c
=> x² + xy + x -y³/3 + 3y = c
which is required solution.
This theorem is also useful in calculation of different types of problems of physics and engineering.
Especially this theorem is used for calculation of potential function in physics.
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