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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 equals 0 is e to the power of integral P d x then P is equal to

  1. fraction numerator 2 x to the power of 2 end exponent minus a x to the power of 3 end exponent over denominator x open parentheses 1 minus x to the power of 2 end exponent close parentheses end fraction    
  2. open parentheses 2 x to the power of 2 end exponent minus 1 close parentheses    
  3. fraction numerator 2 x to the power of 2 end exponent minus 1 over denominator a x to the power of 3 end exponent end fraction    
  4. fraction numerator open parentheses 2 x to the power of 2 end exponent minus 1 close parentheses over denominator x open parentheses 1 minus x to the power of 2 end exponent close parentheses end fraction    

The correct answer is: fraction numerator open parentheses 2 x to the power of 2 end exponent minus 1 close parentheses over denominator x open parentheses 1 minus x to the power of 2 end exponent close parentheses end fraction


    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 equals 0 is e to the power of integral P d x then P is equal to fraction numerator open parentheses 2 x to the power of 2 end exponent minus 1 close parentheses over denominator x open parentheses 1 minus x to the power of 2 end exponent close parentheses end fraction

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