Page 128 - DMTH202_BASIC_MATHEMATICS_II

Unit 9: Solution of Differential Equation

Notes
dy 1  y 2
Task Solve the differential equation: 
dx 1  x 2

9.4.3 Exact Differential Equations

A first order first degree equation of the form
M(x, y)dx + N(x, y) dy = 0

is said to be exact if its left hand side quantity is the exact differential of some function u(x, y).
Thus for the above differential equation to be exact,
du Mdx + Ndy = 0.

The solution of the above equation is given by
u(x, y) = C.

Did u know? Necessary and sufficient condition for a differential equation of the form
 M  N
M( x, y)dx + N(x, y)dy to be exact is  .
y  x 

Working Rule for Solving Exact Differential Equation

The solution of an exact differential equation Mdx + Ndy = 0 is

 Mdx   (terms of N that do not contain )dy  C
x
y constant


Example: Solve  (xy) (dx dy) = dx + dy
Solution:

The given equation is

 (xy) (dx dy) = dx + dy
or, (xy   1)dy = 0
  1)dx(xy +

Here, M = x   1y , N  1(xy + 1).
Now,
 N  M
    1.
x y
Therefore, the given equation is exact. It’s solution is given by

 Mdx   (terms of N that do not contain )dy  C
x
y constant

 (x y 1)dx   ( y +1)dy  C

i.e.,
y constant

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