The differential equation $(3 x+4 y+1) d x+(4 x+5 y+1) d y=0$ represents a family of

The given differential equation is $(3 x+4 y+1) d x+(4 x+5 y+1) d y=0$ Comparing eq. (i) with $\mathrm{Mdx}+\mathrm{Ndy}=0$, we get $M=3 x+4 y+1$ and $\mathrm{N}=4 \mathrm{x}+5 \mathrm{y}+1$ Here, $\frac{\partial \mathrm{M}}{\partial \mathrm{y}}=\frac{\partial \mathrm{N}}{\partial \mathrm{x}}=4$ Hence, eq. (i) is exact and solution is given by $\begin{aligned} & \int(3 x+4 y+1) d x+\int(5 y+1) d y=C \\ \Rightarrow & \frac{3 x^{2}}{2}+4 x y+x+\frac{5 y^{2}}{2}+y-C=0 \\ \Rightarrow & 3 x^{2}+8 x y+2 x+5 y^{2}+2 y-2 C=0 \\ \Rightarrow & 3 x^{2}+2.4 x y+2 x+5 y^{2}+2 y+C^{\prime}=0 \\ & \text { where, } C^{\prime}=-2 C \end{aligned}$ On comparing eq. (ii) with standard form of conic section $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+C=0$ We get, $\mathrm{a}=3, \mathrm{~h}=4, \mathrm{~b}=5$ Here, $h^{2}-a b=16-15=1>0$ Hence, the solution of differential equation represents family of hyperbolas.