The given differential equation is
(3x + 4y + 1) dx + (4x + 5y + 1) dy = 0
Comparing eq. (i) with Mdx + Ndy = 0, we get
M = 3x + 4y + 1 and N = 4x + 5y + 1
Here
=
= 4
Hence, eq. (i) is exact and solution is given by
∫ (3x + 4y +1) dx + ∫ (5y +1) dy = C
⇒
+ 4xy + x +
+ y - C = 0
⇒
3x2 + 8xy + 2x +
5y2 + 2y + 2y - 2C = 0
⇒
3x2 + 2.4 xy + 2x +
5y2 + 2y + C' = 0 ... (ii)
where , C' = - 2C
On comparing eq. (ii) with standard form of conic section
ax2 + 2hxy +
by2 + 2gx + 2fy + C = 0
We get
a = 3 , h = 4 , b = 5
Here ,
h2 - ab = 16 - 15 = 1 > 0
Hence, the solution of differential equation represents family of hyperbolas.