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

Solution

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.

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Vector Algebra

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

Some More Questions From Vector Algebra Chapter

Determine order and degree (if defined) of differential equations:
y' + y = e^x

Determine order and degree (if defined) of differential equations:
y'' + (y')² + 2 y = 0

Determine order and degree (if defined) of differential equations:
y″ + 2y′ + sin y = 0

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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Vector Algebra

Solve the following differential equation:

Some More Questions From Vector Algebra Chapter

Determine order and degree (if defined) of differential equations:y' + y = ex

Determine order and degree (if defined) of differential equations:y'' + (y')2 + 2 y = 0

Determine order and degree (if defined) of differential equations:y″ + 2y′ + sin y = 0

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

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

Determine order and degree (if defined) of differential equations:
y' + y = e^x

Determine order and degree (if defined) of differential equations:
y'' + (y')² + 2 y = 0

Determine order and degree (if defined) of differential equations:
y″ + 2y′ + sin y = 0