MCQ
Choose the correct answer from the given four options.The solution of the equation $(2\text{y}-1)\text{dx}-(2\text{x}+3)\text{dy}=0$ is :
  • A
    $\frac{2\text{x}-1}{2\text{y}+3}=\text{k}$
  • B
    $\frac{2\text{y}+1}{2\text{x}-3}=\text{k}$
  • $\frac{2\text{x}+3}{2\text{y}-1}=\text{k}$
  • D
    $\frac{2\text{x}-1}{2\text{y}-1}=\text{k}$

Answer

Correct option: C.
$\frac{2\text{x}+3}{2\text{y}-1}=\text{k}$
Given is, $(2\text{y}-1)\text{dx}-(2\text{x}+3)\text{dy}=0$
$\Rightarrow(2\text{y}-1)\text{dx}=(2\text{x}+3)\text{dy}$
$\Rightarrow\frac{\text{dx}}{2\text{x}+3}=\frac{\text{dy}}{2\text{y}-1}$
On integrating both sides, we get
$\frac{1}{2}\log(2\text{x}+3)=\frac{1}{2}\log(2\text{y}-1)+\log\text{C}$
$\Rightarrow\frac{1}{2}\log(2\text{x}+3)-\log(2\text{y}-1)=\log\text{C}$
$\Rightarrow\frac{1}{2}\log\Big(\frac{2\text{x}+3}{2\text{y}-1}\Big)=\text{C}^2$
$\Rightarrow\Big(\frac{2\text{x}+3}{2\text{y}-1}\Big)^{\frac{1}{2}}=\text{C}$
$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{C}^2$
$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{C}^2$
$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{k},$ Where $ \text{k}=\text{c}^2$

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