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  • The Integrating Factor of the differential equation ( 1 minus y to the power 2) dy by dx plus y x equals ay ( minus 1 less than y less than 1) is (A) (B) (C) (D)

Q19.    The Integrating Factor of the differential equation    (1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1) is

            (A)    \frac{1}{{y^2 -1}}

            (B)    \frac{1}{\sqrt{y^2 -1}}

            (C)    \frac{1}{{1 - y^2 }}

            (D)    \frac{1}{\sqrt{1 - y^2 }}    

Answers (1)

best_answer

Given equation is
(1 - y^2)\frac{dx}{dy} + yx = ay \ \ (-1<y<1)
we can rewrite it as
\frac{dx}{dy}+\frac{yx}{1-y^2}= \frac{ay}{1-y^2}
It is     \frac{dx}{dy}+px= Q   type of equation where p = \frac{y}{1-y^2}\ and \ Q = \frac{ay}{1-y^2}
Now,
I.F. = e^{\int pdy}= e^{\int \frac{y}{1-y^2}dy}= e^{\frac{\log |1 - y^2|}{-2}}= (1-y^2)^{\frac{-1}{2}}= \frac{1}{\sqrt{1-y^2}}
Therefore, the correct answer is (D)

Posted by

Gautam harsolia

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