Answer:
Option D
Explanation:
The given differential equation
(3x+4y+1)dx+(4x+5y+1)dy=0 .........(i)
comparing eq.(i) with Mdx+Ndy=0, we get
M=3x+4y+1
and N= 4x+5y+1
Here, $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial y}=4$
Hence, eq.(i) is exact and solution is given by
$\int (3x+4y+1)dx+\int(5y+1)dy= C$
$\Rightarrow \frac{3x^{2}}{2}+4xy+x+\frac{5y^{2}}{2}+y-C=0$
$\Rightarrow 3x^{2}+8xy+2x+5y^{2}+2y-2C=0$
$\Rightarrow 3x^{2}+2.4xy+2x+5y^{2}+2y+C'=0$ ......(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
Herece , the solution of differential equation represents family of hyperbolas.
