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Differential Equations

Short Answer Type

111. Solve the following differential equation:

open parentheses straight x squared straight y plus straight x squared close parentheses space dx space plus space left parenthesis straight x space straight y squared minus straight y squared right parenthesis space dy space equals space 0

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112. Solve the following differential equation:
$straight y left parenthesis 1 minus straight x squared right parenthesis dy over dx space equals space straight x left parenthesis 1 plus straight y squared right parenthesis$

The given differential equation is
$straight y left parenthesis 1 minus straight x squared right parenthesis dy over dx space equals space straight x left parenthesis 1 plus straight y squared right parenthesis$
Separating the variables, we get,
$fraction numerator straight y over denominator 1 plus straight y squared end fraction dy space equals space fraction numerator straight x over denominator 1 minus straight x squared end fraction dx$
Integrating, $integral fraction numerator 2 straight y over denominator 1 plus straight y squared end fraction dy space equals space minus integral fraction numerator negative 2 straight x over denominator 1 minus straight x squared end fraction dx$
$therefore space space space log space space open vertical bar 1 plus straight y squared close vertical bar space equals space minus log space open vertical bar 1 minus straight x squared close vertical bar plus log space straight c therefore space space log space open vertical bar 1 minus straight x squared close vertical bar plus log open vertical bar 1 plus straight y squared close vertical bar space equals space log space straight c therefore space space log space open vertical bar 1 minus straight x squared close vertical bar plus log space open vertical bar 1 plus straight y squared close vertical bar space equals space log space straight c therefore space log space open vertical bar left parenthesis 1 minus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis close vertical bar space equals space log space straight c therefore space space space open vertical bar left parenthesis 1 minus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis close vertical bar space equals space straight c therefore space space left parenthesis 1 minus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis space equals space straight A comma space where space straight A space is space constant.$
This is required solution.