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Vector Algebra

Question
CBSEENMA12032996

Find a one-parameter family of solutions of the following differential equations, indicating carefully the interval in which the solutions are valid
dy over dx equals fraction numerator straight x plus 1 over denominator 2 minus straight y end fraction comma space space space space straight y space not equal to 2

Solution
The given differential equation is
                         dy over dx equals fraction numerator straight x plus 1 over denominator 2 minus straight y end fraction
Separating the variables, we get,
                        left parenthesis 2 minus straight y right parenthesis space dy space equals space left parenthesis straight x plus 1 right parenthesis space dx
Integrating,  integral left parenthesis 2 minus straight y right parenthesis space dy space equals space left parenthesis straight x plus 1 right parenthesis space dx
therefore space space space space space space space 2 straight y minus straight y squared over 2 space equals space straight x squared over 2 plus straight x plus straight c apostrophe
or                  4 straight y minus straight y squared space equals space straight x squared plus 2 straight x plus straight c space space space space space space where space straight c space equals space 2 straight c apostrophe
or              straight x squared plus straight y squared plus 2 straight x minus 4 straight y plus straight c space equals space 0
which is required solution.