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Vector Algebra

Question
CBSEENMA12033070

Find the particular solution of the differential equation dy over dx space equals space minus 4 xy squared given that y = 1,  when x = 0.

Solution
The given differential equation is
                                         dy over dx equals negative 4 space straight x space straight y squared
Separating the variables, we get,
                     1 over straight y squared dy space equals space minus 4 straight x space dx
Integrating,   integral 1 over straight y squared dy space equals space minus 4 integral space straight x space dx space space or space space space integral space straight y to the power of negative 2 end exponent space dy space equals space minus 4 space integral space straight x space dx
therefore space space space space space space space space space space space space fraction numerator straight y to the power of negative 1 end exponent over denominator negative 1 end fraction space equals negative 4 straight x squared over 2 plus straight c
or space space space space space space space space space space space space space space space space fraction numerator negative 1 over denominator straight y end fraction space equals space minus 2 straight x squared plus straight c space space or space space space 1 over straight y space equals space 2 straight x squared minus straight c
or space space space space space space space space space space space space space space space space straight y space equals space fraction numerator 1 over denominator 2 straight x squared minus straight c end fraction space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis
Now y = 1 when x = 0
therefore space space space space space space space space space space space space space space space space space space space space space space 1 space equals space fraction numerator 1 over denominator 0 minus straight c end fraction space space or space space space 1 space equals space 1 over straight c space space rightwards double arrow space space straight c space equals space 1
Putting c = -1 in (1), we get,
                                          straight y space equals fraction numerator 1 over denominator 2 straight x squared plus 1 end fraction comma space which space is space required space solution. space

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