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Differential Equations

Question
CBSEENMA12033208

For the given differential equation, find a particular solution satisfying the given condition:
   left parenthesis 1 plus straight x squared right parenthesis space dy over dx space plus space 2 space straight x space straight y space space equals space fraction numerator 1 over denominator 1 plus straight x squared end fraction semicolon space straight y space equals space 0 space space when space straight x space equals space 1


                     

Solution
The given differential equation is
                 left parenthesis 1 plus straight x squared right parenthesis space dy over dx plus 2 xy space equals space fraction numerator 1 over denominator 1 plus straight x squared end fraction
or    dy over dx plus fraction numerator 2 straight x over denominator 1 plus straight x squared end fraction straight y space equals space fraction numerator 1 over denominator left parenthesis 1 plus straight x squared right parenthesis squared end fraction
Comparing it with  dy over dx plus straight P space straight y space equals space straight Q comma space we space get comma space straight P space equals space fraction numerator 2 straight x over denominator straight x squared plus 1 end fraction comma space space straight Q space equals space fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis squared end fraction
       integral straight P space dx space equals space integral fraction numerator 2 straight x over denominator straight x squared plus 1 end fraction dx space equals space log space left parenthesis straight x squared plus 1 right parenthesis comma space space straight e to the power of integral straight P space dx end exponent space equals straight e to the power of log space left parenthesis straight x squared plus 1 right parenthesis end exponent space equals straight x squared plus 1
therefore solution of differential equation is
                             straight y. space straight e to the power of integral straight P space dx end exponent space equals integral straight Q. space straight e to the power of integral straight P space dx end exponent space dx space plus space straight c
or                  straight y. left parenthesis straight x squared plus 1 right parenthesis space equals integral fraction numerator 1 over denominator left parenthesis straight x squared plus 1 right parenthesis squared end fraction. space left parenthesis straight x squared plus 1 right parenthesis space dx space plus space straight c
or             straight y left parenthesis straight x squared plus 1 right parenthesis space equals space integral fraction numerator 1 over denominator straight x squared plus 1 end fraction dx plus straight c
or           straight y left parenthesis straight x squared plus 1 right parenthesis space equals space tan to the power of negative 1 end exponent straight x plus straight c
or                                  straight y equals fraction numerator tan to the power of negative 1 end exponent straight x over denominator straight x squared plus 1 end fraction plus fraction numerator straight c over denominator straight x squared plus 1 end fraction                                 ...(1)
Now y = 0  when x = 1
therefore space space space space 0 space equals space fraction numerator tan to the power of negative 1 end exponent 1 over denominator 1 plus 1 end fraction plus fraction numerator straight c over denominator 1 plus 1 end fraction space space space rightwards double arrow space space space 0 space equals space straight pi over 4 plus straight c space space space rightwards double arrow space space space straight c space equals space minus straight pi over 4
therefore space space space space from space left parenthesis 1 right parenthesis comma space space straight y space equals space fraction numerator tan to the power of negative 1 end exponent straight x over denominator straight x squared plus 1 end fraction minus fraction numerator straight pi over denominator 4 left parenthesis straight x squared plus 1 right parenthesis end fraction
which is required solution.                 

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