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
δM
δy
=
δN
δx
= 4 Hence, eq. (i) is exact and solution is given by ∫ (3x + 4y +1) dx + ∫ (5y +1) dy = C ⇒
3x2
2
+ 4xy + x +
5y2
2
+ 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.