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The Integrating Factor of the Differential Equation ( 1 − Y 2 ) D X D Y + Y X = a Y ( − 1 < Y < 1 ) is - Mathematics

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Question

The integrating factor of the differential equation \[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\left( - 1 < y < 1 \right)\] is ______.

Options

  • \[\frac{1}{y^2 - 1}\]

  • \[\frac{1}{\sqrt{y^2 - 1}}\]

  • \[\frac{1}{1 - y^2}\]

  • \[\frac{1}{\sqrt{1 - y^2}}\]

MCQ
Fill in the Blanks
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Solution

The integrating factor of the differential equation \[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\left( - 1 < y < 1 \right)\] is `bbunderline(1/(sqrt(1 - "y"^2)))`.
Explanation:
We have,
\[\left( 1 - y^2 \right)\frac{dx}{dy} + yx = ay\]
\[\frac{dx}{dy} + \frac{y}{1 - y^2} x = \frac{ay}{1 - y^2}\]
\[\text{Comparing with }\frac{dx}{dy} + Px = Q,\text{ we get }\]
\[P = \frac{y}{1 - y^2} \]
\[Q = \frac{ay}{1 - y^2}\]
Now, 
\[ I . F . = e^{\int\frac{y}{1 - y^2}dy} \]
\[ = e^{- \frac{1}{2}\int\frac{- 2y}{1 - y^2}dy} \]
\[ = e^{- \frac{1}{2}\log\left| 1 - y^2 \right|} \]
\[ = e^{log\left| \frac{1}{\sqrt{1 - y^2}} \right|} \]
\[ = \frac{1}{\sqrt{1 - y^2}}\]
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Chapter 22: Differential Equations - MCQ [Page 144]

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