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Vector Algebra

Question
CBSEENMA12033000

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 equals square root of 4 minus straight x squared end root comma space space minus 2 less than straight y less than 2


Solution
The given differential equation is
                        dy over dx equals square root of 4 minus straight y squared end root
Separating the variables, we get,
                     fraction numerator 1 over denominator square root of 4 minus straight y squared end root end fraction dy space equals space dx
Integrating,    integral fraction numerator 1 over denominator square root of 4 minus straight y squared end root end fraction dy space equals space integral 1 space dx
therefore space space space integral fraction numerator 1 over denominator square root of left parenthesis 2 right parenthesis squared minus straight y squared end root end fraction dy space equals space integral 1 space dx
therefore space space sin to the power of negative 1 end exponent open parentheses straight y over 2 close parentheses space equals space straight x plus straight c
therefore space space space space space space space space space space straight y over 2 space equals space sin space left parenthesis straight x plus straight c right parenthesis
or            straight y equals 2 space sin space left parenthesis straight x plus straight c right parenthesis comma  which is required solution.

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