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Differential Equations

Question
CBSEENMA12033117
Wired Faculty App

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
left parenthesis straight x plus 2 straight y right parenthesis space dx space minus space left parenthesis 2 straight x minus straight y right parenthesis space dy space equals space 0

Solution

The given differential equation is
                    left parenthesis straight x plus 2 straight y right parenthesis space dx space minus space left parenthesis 2 straight x minus straight y right parenthesis space dy space equals space 0
or                 dy over dx space equals space fraction numerator straight x plus 2 straight y over denominator 2 straight x minus straight y end fraction
Put y = v x  so that   dy over dx space equals space straight v plus straight x dv over dx
therefore space space space straight v plus straight x dv over dx space equals fraction numerator straight x plus 2 vx over denominator 2 straight x minus vx end fraction therefore space space space space straight v plus straight x dv over dx space equals space fraction numerator 1 plus 2 straight v over denominator 2 minus straight v end fraction therefore space space space space space straight x dv over dx space equals space fraction numerator 1 plus 2 straight v over denominator 2 minus straight v end fraction minus straight v therefore space space space space straight x dv over dx space equals space fraction numerator 1 plus 2 straight v minus 2 straight v plus straight v squared over denominator 2 minus straight v end fraction therefore space space space space space space straight x dv over dx space equals space fraction numerator 1 plus straight v squared over denominator 2 minus straight v end fraction therefore space space space fraction numerator 2 minus straight v over denominator 1 plus straight v squared end fraction dv space equals space 1 over straight x dx therefore space space space space integral fraction numerator 2 minus straight v over denominator 1 plus straight v squared end fraction dv space equals space integral 1 over straight x dx therefore space space space 2 space integral fraction numerator 1 over denominator 1 plus straight v squared end fraction dv minus 1 half integral fraction numerator 2 straight v over denominator 1 plus straight v squared end fraction dv space equals space integral 1 over straight x dx therefore space 2 space tan to the power of negative 1 end exponent straight v space minus space 1 half log space left parenthesis 1 plus straight v squared right parenthesis space equals space log space open vertical bar straight x close vertical bar plus space log space straight c subscript 1 therefore space space 4 space tan to the power of negative 1 end exponent straight v space minus space log space left parenthesis 1 plus straight v squared right parenthesis space equals space 2 space log space open vertical bar straight x close vertical bar space plus space 2 space log space straight c subscript 1 therefore space space 4 space tan to the power of negative 1 end exponent straight y over straight x minus log space open parentheses 1 plus straight y squared over straight x squared close parentheses space equals space log space straight x squared plus space log space straight c therefore space space space 4 space tan to the power of negative 1 end exponent straight y over straight x minus log space left parenthesis straight x squared plus straight y squared right parenthesis space plus space log space straight x squared space equals space log space straight x squared plus space log space straight c therefore space space space 4 space tan to the power of negative 1 end exponent straight y over straight x minus log space left parenthesis straight x squared plus straight y squared right parenthesis space equals space log space straight c
which is the required solution. 

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