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Differential Equations

Question
CBSEENMA12033113

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(3 x y + y2) dx = (x2 + x y) dy

Solution
The given differential equation is
                    (3 x y + y2) dx = (x2 + x y) dy
or                                    dy over dx equals fraction numerator 3 xy plus straight y squared over denominator straight x squared plus xy end fraction
Put y = v x so that dy over dx equals straight v plus straight x dv over dx
therefore space space space space space space straight v plus straight x dv over dx equals fraction numerator 3 straight x squared straight v plus straight v squared straight x squared over denominator straight x squared plus straight x squared straight v end fraction
therefore space space space straight v plus straight x dv over dx space equals space fraction numerator 3 straight v plus straight v squared over denominator 1 plus straight v end fraction
therefore space space space space space space straight x dv over dx space equals space fraction numerator 3 straight v plus straight v squared over denominator 1 plus straight v end fraction minus straight v space space space space space or space space space straight x dv over dx space equals space fraction numerator 3 straight v plus straight v squared minus straight v minus straight v squared over denominator 1 minus straight v end fraction
therefore space space space space space space straight x dv over dx equals fraction numerator 2 straight v over denominator 1 plus straight v end fraction
Separating the variables, fraction numerator 1 plus straight v over denominator straight v end fraction dv space equals space 2 over straight x dx
Integrating,  integral open parentheses 1 over straight v plus 1 close parentheses space dv space equals space 2 integral 1 over straight x dx
therefore space space space space space log space open vertical bar straight v close vertical bar plus straight v space equals space 2 space log space open vertical bar straight x close vertical bar space plus space straight c
therefore space space space space log space open vertical bar straight y over straight x close vertical bar plus straight y over straight x space equals space 2 space log space open vertical bar straight x close vertical bar plus straight c
is the required solution. 

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