MCQ
The solution of the differential equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{ax}+\text{g}}{\text{by}+\text{f}}$ represents a circle when,
  • A
    $\text{a}=\text{b}$
  • $\text{a}=-\text{b}$
  • C
    $\text{a}=-2\text{b}$
  • D
    $\text{a}=2\text{b}$

Answer

Correct option: B.
$\text{a}=-\text{b}$
We have,
$​​\frac{\text{dy}}{\text{dx}}=\frac{\text{ax}+\text{g}}{\text{by}+\text{f}}$
$\Rightarrow (\text{by}+\text{f})\text{dy}=(\text{ax}+\text{g})\text{dx}$
Intergrating both sides, we get
$\Rightarrow \int(\text{by}+\text{f})\text{dy}=\int(\text{ax}+\text{g})\text{dx}$
$\Rightarrow \text{b}\frac{\text{y}^{2}}{2}+\text{fy}=\text{a}\frac{\text{x}^{2}}{2}+\text{gx}+\text{C}$
$\Rightarrow \text{b}\frac{\text{y}^{2}}{2}+\text{fy}-\text{a}\frac{\text{x}^{2}}{2}-\text{gx}=\text{C}$
$\Rightarrow \text{b}\text{y}^{2}+2\text{fy}-\text{a}\text{x}^{2}-2\text{gx}-2\text{C}=0$
The above equation resprasents a circle.
Therefore, the coffrcients of $x^2$ and $y^2$ must be equal.
$-\text{a}=\text{b}$
$\Rightarrow \text{a}=-\text{b}$

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