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Differential Equations

Question
CBSEENMA12033131

Find a one parameter family of solutions of each of the following differential equation:
 (x2 + x y) dy = (x2 + y2) dx  

Solution
The given differential equation is
                                      (x2 + x y) dy = (x2 + y2) dx  
or                     dy over dx space equals space fraction numerator straight x squared plus straight y squared over denominator straight x squared plus xy end fraction                                   ...(1)
Put y = v x so that dy over dx equals straight v plus straight x dv over dx
therefore space space space from space left parenthesis 1 right parenthesis comma space space space space straight v plus straight x dv over dx equals fraction numerator straight x squared plus straight v squared space straight x squared over denominator straight x squared plus straight v space straight x squared end fraction
rightwards double arrow space space space space straight v plus straight x dv over dx equals fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction space space space space space space space rightwards double arrow space space space straight x dv over dx space equals fraction numerator 1 plus straight v squared over denominator 1 plus straight v end fraction minus straight v
therefore space space space space straight x dv over dx space equals space fraction numerator 1 minus straight v over denominator 1 plus straight v end fraction space space space space rightwards double arrow space space space space fraction numerator 1 plus straight v over denominator 1 minus straight v end fraction dv space equals space 1 over straight x dx space space space space rightwards double arrow space space integral open parentheses negative 1 plus fraction numerator 2 over denominator 1 minus straight v end fraction close parentheses dv space equals integral 1 over straight x dx
therefore space space minus straight v minus 2 space log space left parenthesis 1 minus straight v right parenthesis space equals space log space straight x space plus straight c space space space space rightwards double arrow space space space space space minus straight y over straight x minus 2 space log space open parentheses 1 minus straight y over straight x close parentheses space equals space log space straight x plus straight c
therefore space space space space space straight y minus 2 straight x space log space open parentheses fraction numerator straight x minus straight y over denominator straight x end fraction close parentheses space equals space straight x space logx plus space cx
which is required solution. 

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