If integrating factor of x ( 1 − x 2 ) d y + ( 2 x 2 y − y − a x 3 ) d x = 0 is e ∫ P d x , then P is equal to Options: A . &nbsp2x2−ax3x(1−x2) B . &nbsp(2x2−1) C . &nbsp2x2−1ax3 D . &nbsp(2x2−1)x(1−x2)

Answer: Option D : D x ( 1 − x 2 ) d y + ( 2 x 2 y − y − a x 3 ) d x = 0 d y d x + ( 2 x 2 − 1 ) x ( 1 − x 2 ) y = a x 2 ( 1 − x 2 ) , ∴ P = 2 x 2 − 1 x ( 1 − x 2 ) .

If integrating factor of x(1−x2)dy+(2x2y−y−ax3)dx=0 is e&#x

Question

If integrating factor of

x (1 - x^{2}) d y + (2 x^{2} y - y - a x^{3}) d x = 0

e^{\int P d x}

, then P is equal to

Options:

A . 2x2−ax3x(1−x2)

B . (2x2−1)

C . 2x2−1ax3

D . (2x2−1)x(1−x2)

Answer: Option D
:
D

x (1 - x^{2}) d y + (2 x^{2} y - y - a x^{3}) d x = 0 \frac{d y}{d x} + \frac{(2 x^{2} - 1)}{x (1 - x^{2})} y = \frac{a x^{2}}{(1 - x^{2})}, ∴ P = \frac{2 x^{2} - 1}{x (1 - x^{2})}

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