Find a one-parameter family of solutions of the following differential equation, indicating carefully the interval in which the solutions are valid:
$dy over dx space equals space fraction numerator 1 plus straight y squared over denominator 1 plus straight x squared end fraction$

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Solution

The given differential equation is

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

Separating the variables, we get,

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

Integrating,

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

therefore space space space tan to the power of negative 1 end exponent straight y space equals space tan to the power of negative 1 end exponent straight x plus straight c

is the solution.

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