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CBSE Class 12 Math 2023 All Sets Solved Paper

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Question : 3 of 20
Marks: +1, -0
The integrating factor of the differential equation (1−y2)dxdy+yx=ay,(1-y^2)\frac{dx}{dy} + yx = ay, (-1 < y < 1) is
Solution:  
Given differential equation is (1−y2)dxdy+yx=ay,(−1<y<1)(1-y^2)\frac{dx}{dy} + yx = ay,(-1 < y < 1)
⇒dxdy+yx1−y2=ay1−y2\Rightarrow \frac{dx}{dy} + \frac{yx}{1-y^2} = \frac{ay}{1-y^2}
⇒dxdy+(y1−y2)x=ay1−y2\Rightarrow \frac{dx}{dy} + \left(\frac{y}{1-y^2}\right)x = \frac{ay}{1-y^2}
The given equation in form of dxdy+px=q\frac{dx}{dy} + px = q
∴I.F.=e∫p dy\therefore \text{I.F.} = e^{\int p \, dy}
=e∫y1−y2 dy= e^{\int \frac{y}{1-y^2} \, dy}
=e∫−12(−2y1−y2) dy= e^{\int -\frac{1}{2} \left(\frac{-2y}{1-y^2}\right) \, dy}
=e−12log⁡(1−y2)= e^{-\frac{1}{2} \log(1-y^2)}
=elog⁡11−y2= e^{\log \frac{1}{\sqrt{1-y^2}}}
∴I.F.=11−y2\therefore \text{I.F.} = \frac{1}{\sqrt{1-y^2}}
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