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Differential Equations

Question
CBSEENMA12033140

For the given differential equations, find the particular solution satisfying the given condition:
left parenthesis straight x plus straight y right parenthesis space dy space plus space left parenthesis straight x minus straight y right parenthesis dx space equals space 0 space semicolon space space straight y space equals space 1 space space space when space straight x space equals space 1

Solution
The given differential equation is
             left parenthesis straight x plus straight y right parenthesis space dy space plus space left parenthesis straight x minus straight y right parenthesis space dx space equals space 0 comma space space space or space space space left parenthesis straight x plus straight y right parenthesis space dy space equals space minus left parenthesis straight x minus straight y right parenthesis space dx
therefore space space space space space dy over dx space equals space minus fraction numerator straight x minus straight y over denominator straight x plus straight y end fraction space space space or space space space dy over dx space equals space fraction numerator straight y minus straight x over denominator straight y plus straight x end fraction                        ...(1)
Put y = v  x,  so that dy over dx space equals straight v plus straight x dv over dx
therefore space space space space space from space left parenthesis 1 right parenthesis comma space space space space straight v plus straight x dv over dx space equals space fraction numerator straight v space straight x space minus straight x over denominator vx plus straight x end fraction space space or space space straight v plus straight x dv over dx space equals space fraction numerator straight v minus 1 over denominator straight v plus 1 end fraction
or       straight x dy over dx space equals fraction numerator straight v minus 1 over denominator straight v plus 1 end fraction minus straight v space equals space fraction numerator negative 1 minus straight v squared over denominator straight v plus 1 end fraction space equals space fraction numerator negative left parenthesis 1 plus straight v squared right parenthesis over denominator 1 plus straight v end fraction space space or space space fraction numerator 1 plus straight v over denominator 1 plus straight v squared end fraction dv space equals negative dx over straight x
or      integral fraction numerator 1 plus straight v over denominator 1 plus straight v squared end fraction dv space equals space minus integral dx over straight x
or     integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv plus 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space minus integral 1 over straight x dx
or               tan to the power of negative 1 end exponent straight v space plus space 1 half log space open vertical bar 1 plus straight v squared close vertical bar plus log space open vertical bar straight x close vertical bar space equals space straight c
or               tan to the power of negative 1 end exponent straight v plus 1 half left square bracket log space open vertical bar 1 plus straight v squared close vertical bar space plus space 2 space log space open vertical bar straight x close vertical bar right square bracket space equals space straight c
or           tan to the power of negative 1 end exponent straight v plus 1 half log space open vertical bar straight x squared close vertical bar space space open vertical bar 1 plus straight v squared close vertical bar space equals space straight c
or space space tan space to the power of negative 1 end exponent straight y over straight x space plus space 1 half space log space open vertical bar straight x squared plus open parentheses 1 plus straight y squared over straight x squared close parentheses close vertical bar space equals space straight c
or space space space tan space to the power of negative 1 end exponent straight y over straight x plus 1 half log space open vertical bar straight x squared plus straight y squared close vertical bar space equals space straight c
therefore space space space space tan to the power of negative 1 end exponent straight y over straight x plus 1 half log space left parenthesis straight x squared plus straight y squared right parenthesis space equals space straight c space space space space space space space space space space space space space space space space space space... left parenthesis 2 right parenthesis
                 Now y = 1  when x = 1
therefore space space space space space space tan to the power of negative 1 end exponent 1 plus 1 half log 2 space equals space straight c space space space space space rightwards double arrow space space space space space straight c space equals space straight pi over 4 plus 1 half log space 2
therefore space space space space from space left parenthesis 2 right parenthesis comma space space tan to the power of negative 1 end exponent straight y over straight x plus 1 half log space left parenthesis straight x squared plus straight y squared right parenthesis space equals space straight pi over 4 plus 1 half log space 2
                              is the required solution. 

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