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Differential Equations

Question
CBSEENMA12033111

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(x - y) y' = x + 2 y 

Solution
The given differential equation is
       left parenthesis straight x minus straight y right parenthesis space dy over dx space equals space straight x plus 2 straight y
or                  dy over dx equals fraction numerator straight x plus 2 straight y over denominator straight x minus straight y end fraction                                     ...(1)
Put y = vx 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 straight v plus straight x dv over dx equals fraction numerator straight x plus 2 space straight v space straight x over denominator 1 minus straight v space straight x end fraction space space space space space space or space space space space space space straight v plus straight x dv over dx equals fraction numerator 1 plus 2 straight v over denominator 1 minus straight v end fraction
therefore space space space space space space space space space straight x dv over dx equals fraction numerator 1 plus 2 straight v over denominator 1 minus straight v end fraction minus straight v space space space space space space space space space space space space space space space space space space space space space space rightwards double arrow space space space straight x space dv over dx space equals fraction numerator 1 plus 2 space straight v minus straight v plus straight v squared over denominator 1 minus straight v end fraction
or space space space space space space space space straight x dv over dx equals fraction numerator 1 plus straight v plus straight v squared over denominator 1 minus straight v end fraction
Separating the variables,  fraction numerator 1 minus straight v over denominator 1 plus straight v plus straight v squared end fraction dv space equals space 1 over straight x dx
Integrating,      integral fraction numerator 1 minus straight v over denominator 1 plus straight v plus straight v squared end fraction dv space equals space integral 1 over straight x dx
therefore space space space space integral fraction numerator negative begin display style 1 half end style left parenthesis 2 straight v plus 1 right parenthesis plus begin display style 3 over 2 end style over denominator 1 plus straight v plus straight v squared end fraction dv space equals space integral 1 over straight x dv
therefore space space space minus 1 half integral fraction numerator 2 straight v plus 1 over denominator 1 plus straight v plus straight v squared end fraction dv plus 3 over 2 integral fraction numerator 1 over denominator 1 plus straight v plus straight v squared end fraction dv space equals space integral 1 over straight x dx
therefore space space minus 1 half integral fraction numerator 2 straight v plus 1 over denominator 1 plus straight v plus straight v squared end fraction dv plus 3 over 2 integral fraction numerator 1 over denominator open parentheses straight v plus begin display style 1 half end style close parentheses squared plus open parentheses begin display style fraction numerator square root of 3 over denominator 2 end fraction end style close parentheses squared end fraction dv space equals space integral 1 over straight x dx
therefore space space space space minus 1 half log space open vertical bar 1 plus straight v plus straight v squared close vertical bar plus 3 over 2. space fraction numerator 1 over denominator begin display style fraction numerator square root of 3 over denominator 2 end fraction end style end fraction space tan to the power of negative 1 end exponent open parentheses fraction numerator straight v plus begin display style 1 half end style over denominator begin display style fraction numerator square root of 3 over denominator 2 end fraction end style end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar plus straight c subscript 1
therefore space space space minus 1 half log space open vertical bar 1 plus straight y over straight x plus straight y squared over straight x squared close vertical bar plus square root of 3 space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 straight v plus 1 over denominator square root of 3 end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar space plus space straight c subscript 1
therefore space space space space minus 1 half log space open vertical bar fraction numerator straight x squared plus straight x space straight y plus straight y squared over denominator straight x squared end fraction close vertical bar plus square root of 3 space tan to the power of negative 1 end exponent space open parentheses fraction numerator 2 begin display style straight y over straight x end style plus 1 over denominator square root of 3 end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar space plus space straight c subscript 1
therefore space space space space 2 square root of 3 space tan to the power of negative 1 end exponent open parentheses fraction numerator 2 straight y plus straight x over denominator straight x square root of 3 end fraction close parentheses space minus space log space open vertical bar straight x squared plus xy plus straight y squared close vertical bar plus log space straight x squared space equals space log space straight x squared plus straight c
therefore space space space 2 square root of 3 space tan to the power of negative 1 end exponent space open parentheses fraction numerator straight x plus 2 straight y over denominator straight x square root of 3 end fraction close parentheses space minus space log space open vertical bar straight x squared plus straight x space straight y space plus straight y squared close vertical bar space equals space straight c
Which is required solution. 

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