Find a one-parameter family of solutions of the following differential equation, indicating carefully the interval in which the solutions are valid:
$straight y apostrophe space equals left parenthesis 1 plus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis$

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Solution

The given differential equation is

straight y apostrophe space equals left parenthesis 1 plus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis space space space space or space space space space dy over dx equals left parenthesis 1 plus straight x squared right parenthesis thin space left parenthesis 1 plus straight y squared right parenthesis

Separating the variables, we get,

fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space left parenthesis 1 plus straight x squared right parenthesis space dx

Integrating,

integral fraction numerator 1 over denominator 1 plus straight y squared end fraction dy space equals space integral left parenthesis 1 plus straight x squared right parenthesis space dx

therefore space space space space space space space tan to the power of negative 1 end exponent straight y space equals space straight x plus straight x cubed over 3 plus straight c comma space space straight x space element of space straight R space is space the space solution. space

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