Question

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

• A. circles
• B. parabolas
• C. ellipses
• D. hyperbolas

Answer 1

Answer: (D)

The given differential equation is
$(3\,x + 4\,y + 1) d\,x + (4\,x + 5\,y + 1) dy = 0$ ....(i)
Comparing eq. (i) with $Mdx + Ndy = 0$,
we get
$M = 3\,x + 4\,y + 1$
and $N = 4\,x + 5\,y + 1$
Here, $\frac{\partial M}{ \partial y} = \frac{\partial N}{ \partial x} = 4 $
Hence, eq. (i) is exact and solution is given by
$\int \left(3x + 4y +1\right)dx+\int\left(5y + 1\right)dy = C$
$ \Rightarrow \frac{3x^{2}}{2} + 4\,xy +x+\frac{5y^{2}}{2} + y- C = 0 $
$\Rightarrow 3x^{2} + 8\,xy +2\,x +5y^{2} + 2\,y - C = 0$
$ \Rightarrow 3x^{2} + 2.4xy+2\,x+5y^{2} + 2\,y +C' = 0 $ ....(ii)
where, $C' = -2C$
On comparing eq. (ii) with standard form of conic section
$ax^2 + 2\,hxy + by^2 + 2\,gx + 2fy + C = 0$
We get,
$a = 3, h = 4, b = 5$
Here, $h^2 - a\,b = 16 - 1\,5 = 1 > 0$
Hence, the solution of differential equation represents family of hyperbolas.