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Question: The solution of the differential equation \(\left( 1 - x ^ { 2 } \right) ( 1 - y ) d x = x y ( 1 + ...

The solution of the differential equation

(1−x2)(1−y)dx=xy(1+y)dy\left( 1 - x ^ { 2 } \right) ( 1 - y ) d x = x y ( 1 + y ) d y is

A

log⁡[x(1−y)2]=x22+y22−2y+c\log \left[ x ( 1 - y ) ^ { 2 } \right] = \frac { x ^ { 2 } } { 2 } + \frac { y ^ { 2 } } { 2 } - 2 y + c

B

log⁡[x(1−y)2]=x22−y22+2y+c\log \left[ x ( 1 - y ) ^ { 2 } \right] = \frac { x ^ { 2 } } { 2 } - \frac { y ^ { 2 } } { 2 } + 2 y + c

C

log⁡[x(1+y)2]=x22+y22+2y+c\log \left[ x ( 1 + y ) ^ { 2 } \right] = \frac { x ^ { 2 } } { 2 } + \frac { y ^ { 2 } } { 2 } + 2 y + c

D

log⁡[x(1−y)2]=x22−y22−2y+c\log \left[ x ( 1 - y ) ^ { 2 } \right] = \frac { x ^ { 2 } } { 2 } - \frac { y ^ { 2 } } { 2 } - 2 y + c

Answer

log⁡[x(1−y)2]=x22−y22−2y+c\log \left[ x ( 1 - y ) ^ { 2 } \right] = \frac { x ^ { 2 } } { 2 } - \frac { y ^ { 2 } } { 2 } - 2 y + c

Explanation

Solution

(1−x2)(1−y)dx=xy(1+y)dy\left( 1 - x ^ { 2 } \right) ( 1 - y ) d x = x y ( 1 + y ) d y

⇒ ∫y(1+y)(1−y)dy=∫(1−x2)xdx\int \frac { y ( 1 + y ) } { ( 1 - y ) } d y = \int \frac { \left( 1 - x ^ { 2 } \right) } { x } d x;

Now integrate it