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

Solution of the equation (1−x2)dy+xydx=xy2dx\left( 1 - x ^ { 2 } \right) d y + x y d x = x y ^ { 2 } d x is

A

(y−1)2(1−x2)=0( y - 1 ) ^ { 2 } \left( 1 - x ^ { 2 } \right) = 0

B

(y−1)2(1−x)2=c2y2( y - 1 ) ^ { 2 } ( 1 - x ) ^ { 2 } = c ^ { 2 } y ^ { 2 }

C

(y−1)2(1+x2)=c2y2( y - 1 ) ^ { 2 } \left( 1 + x ^ { 2 } \right) = c ^ { 2 } y ^ { 2 }

D

None of these

Answer

(y−1)2(1−x)2=c2y2( y - 1 ) ^ { 2 } ( 1 - x ) ^ { 2 } = c ^ { 2 } y ^ { 2 }

Explanation

Solution

(1−x2)dy+xydx=xy2dx\left( 1 - x ^ { 2 } \right) d y + x y d x = x y ^ { 2 } d x

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

Now on integrating both sides, we get

log⁡(y−1)−log⁡y=−12log⁡(1−x2)+log⁡c\log ( y - 1 ) - \log y = - \frac { 1 } { 2 } \log \left( 1 - x ^ { 2 } \right) + \log c

or 2log⁡(y−1)+log⁡(1−x2)=log⁡y2c22 \log ( y - 1 ) + \log \left( 1 - x ^ { 2 } \right) = \log y ^ { 2 } c ^ { 2 }

Hence the solution is (y−1)2(1−x2)=c2y2( y - 1 ) ^ { 2 } \left( 1 - x ^ { 2 } \right) = c ^ { 2 } y ^ { 2 }.