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Maths-

General

Easy

Question

If integrating factor of $x open parentheses 1 minus x squared close parentheses d y plus open parentheses 2 x squared y minus y minus a x cubed close parentheses d x$ is $e to the power of JPdx$ then P is

$\frac{2 x^{2} - a x^{3}}{x (1 - x^{2})}$
$2 x^{3} - 1$
$\frac{2 x^{2} - 1}{a x^{3}}$
$\frac{2 x^{2} - 1}{x (1 - x^{2})}$

hint Hint:

In this question we have to find the integrating factor of the given function. For this we will simplify the equation to bring it in the form dy/dx+Py=Q, where P and Q are the constants or functions of x. Then we can find the integrating factor using $e to the power of JPdx$ .

The correct answer is: $fraction numerator 2 x squared minus 1 over denominator x open parentheses 1 minus x squared close parentheses end fraction$

$integral x open parentheses 1 minus x squared close parentheses d y plus open parentheses 2 x squared y minus y minus a x cubed close parentheses d x equals 0 rightwards double arrow x open parentheses 1 minus x squared close parentheses fraction numerator d y over denominator d x end fraction plus 2 x squared y minus y minus a x cubed equals 0 rightwards double arrow x open parentheses 1 minus x squared close parentheses fraction numerator d y over denominator d x end fraction plus y open parentheses 2 x squared minus 1 close parentheses equals a x cubed rightwards double arrow fraction numerator d y over denominator d x end fraction plus fraction numerator y open parentheses 2 x squared minus 1 close parentheses over denominator x open parentheses 1 minus x squared close parentheses end fraction equals fraction numerator a x cubed over denominator x open parentheses 1 minus x squared close parentheses end fraction w h i c h space i s space o f space t h e space f o r m space fraction numerator d y over denominator d x end fraction plus P y equals Q t h e space i n e t g a r t i n g space f a c t o r space i s space e to the power of integral P space d x end exponent h e r e comma space P equals fraction numerator 2 x squared minus 1 over denominator x open parentheses 1 minus x squared close parentheses end fraction$

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