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DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS1 Mark
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

$\frac{1}{2}\log(2\text{x}+3)=\frac{1}{2}\log(2\text{y}-1)+\log\text{C}$

$\Rightarrow\frac{1}{2}\log(2\text{x}+3)-\log(2\text{y}-1)=\log\text{C}$

$\Rightarrow\frac{1}{2}\log\Big(\frac{2\text{x}+3}{2\text{y}-1}\Big)=\text{C}^2$

$\Rightarrow\Big(\frac{2\text{x}+3}{2\text{y}-1}\Big)^{\frac{1}{2}}=\text{C}$

$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{C}^2$

$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{C}^2$

$\Rightarrow\frac{2\text{x}+3}{2\text{y}-1}=\text{k},\text{Where}\ \text{k}=\text{c}^2$

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