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Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations1 Mark
MCQ
The integrating factor of the differential equation$(1-\text{y}^{2})\frac{\text{dx}}{\text{dy}}+\text{yx}=\text{ay}(-1<\text{y}<1)$ is:
  • A
    $\frac{1}{\text{y}^{2}-1}$
  • B
    $\frac{1}{\sqrt{\text{y}^{2}+1}}$
  • C
    $\frac{1}{1-\text{y}^{2}}$
  • ✓
    $\frac{1}{\sqrt{1-\text{y}^{3}}}$
✓

Answer

Correct option: D.
$\frac{1}{\sqrt{1-\text{y}^{3}}}$
We have,
$(1-\text{y}^{2})\frac{\text{dx}}{\text{dy}}+\text{yx}=\text{ay}$
$\frac{\text{dx}}{\text{dy}}+\frac{\text{y}}{1-\text{y}^{2}}\text{x}=\frac{\text{ay}}{1-\text{y}^{2}}$
Comparing with we get,
$\text{P}=\frac{\text{y}}{1-\text{y}^{2}}, \text{Q}=\frac{\text{ay}}{1-\text{y}^{2}}$
Now,
$\text{I.F}=\text{e}^{\int\frac{\text{y}}{1-\text{y}^{2}}\text{dy}}$
$=\text{e}^{-\frac{1}{2}\int\frac{-2\text{y}}{1-\text{y}^{2}}\text{dy}}$
$=\text{e}^{-\frac{1}{2}\log|1-\text{y}^{2}|}$
$=\text{e}^{\log\Big|\frac{1}{\sqrt{1-\text{y}^{2}}}\Big|}$
$=\frac{1}{\sqrt{1-\text{y}^{2}}}$

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