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

The solution of the differential equation

(x−y2x)dx=(y−x2y)dy\left( x - y ^ { 2 } x \right) d x = \left( y - x ^ { 2 } y \right) d y is

A

(1−y2)=c2(1−x2)\left( 1 - y ^ { 2 } \right) = c ^ { 2 } \left( 1 - x ^ { 2 } \right)

B

(1+y2)=c2(1−x2)\left( 1 + y ^ { 2 } \right) = c ^ { 2 } \left( 1 - x ^ { 2 } \right)

C

(1+y2)=c2(1+x2)\left( 1 + y ^ { 2 } \right) = c ^ { 2 } \left( 1 + x ^ { 2 } \right)

D

None of these

Answer

(1−y2)=c2(1−x2)\left( 1 - y ^ { 2 } \right) = c ^ { 2 } \left( 1 - x ^ { 2 } \right)

Explanation

Solution

Given equation can be written as

x1−x2dx=y1−y2dy\frac { x } { 1 - x ^ { 2 } } d x = \frac { y } { 1 - y ^ { 2 } } d y

On integrating we get

−12log⁡(1−x2)=−12log⁡(1−y2)+log⁡c- \frac { 1 } { 2 } \log \left( 1 - x ^ { 2 } \right) = - \frac { 1 } { 2 } \log \left( 1 - y ^ { 2 } \right) + \log c

⇒ log⁡(1−x2)−log⁡(1−y2)=−2log⁡c\log \left( 1 - x ^ { 2 } \right) - \log \left( 1 - y ^ { 2 } \right) = - 2 \log c ⇒ 1−x21−y2=c−2\frac { 1 - x ^ { 2 } } { 1 - y ^ { 2 } } = c ^ { - 2 }

Hence(1−y2)=c2(1−x2)\left( 1 - y ^ { 2 } \right) = c ^ { 2 } \left( 1 - x ^ { 2 } \right)