MCQ
An integrating factor for the differential equation $(1 + {y^2})dx - ({\tan ^{ - 1}}y - x)dy = 0$
- A${\tan ^{ - 1}}y$
- ✓${e^{{{\tan }^{ - 1}}y}}$
- C$\frac{1}{{1 + {y^2}}}$
- D$\frac{1}{{x(1 + {y^2})}}$
==> $\frac{{dy}}{{dx}} = \frac{{1 + {y^2}}}{{{{\tan }^{ - 1}}y - x}}$ ==> $\frac{{dx}}{{dy}} = \frac{{{{\tan }^{ - 1}}y}}{{1 + {y^2}}} - \frac{x}{{1 + {y^2}}}$
==> $\frac{{dx}}{{dy}} + \frac{x}{{1 + {y^2}}} = \frac{{{{\tan }^{ - 1}}y}}{{1 + {y^2}}}$
This is equation of the form $\frac{{dx}}{{dy}} + Px = Q$
So, $I.F.$ $ = {e^{\int {P\,dy} }} = {e^{\int {\frac{1}{{1 + {y^2}}}.dy} }} = {e^{{{\tan }^{ - 1}}y}}$.
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