Question

Find the general solution of the differential equation


`(dy)/(dx) = \frac{ {1 + y}^(2) }{ {1 + x}^(2) }`
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Answer 1

`{ \blue{ \tt{ { \tan }^( - 1) y - { \tan}^( - 1) x = c}}}`

Explanation:

This can be written as,

`{ \red{ \tt{ \frac{dy}{ {1 + y}^(2) } = \frac{dx}{1 + {x}^(2) }}}}`

Integrate on both sides

`{ \red{ \tt{∫ \frac{ 1}{1 + {y}^(2)}}}}{ \red{ \tt{dy}}} \: = { \red{ \tt{∫ \frac{1}{1 + {x}^(2) }}}}{ \red{ \tt{dx}}}`

`{ \red{ \tt{ { \tan}^( - 1) y = { \tan }^( - 1) x + c}}}`

`{ \red{ \tt{ { \tan }^( - 1)y - { \tan }^( - 1)x = c}}}`