Question

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
2 x y dx + (x² + 2 y²) dy = 0

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Solution

The given differential equation is
2 x y dx + (x² + 2 y²) dy = 0 or

left parenthesis straight x squared plus 2 straight y squared right parenthesis space dy space equals space minus 2 xy space dx

therefore space space space space dy over dx space equals space minus fraction numerator 2 xy over denominator straight x squared plus 2 straight y squared 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 straight v plus straight x dv over dx equals negative fraction numerator 2. straight x. space straight v space straight x over denominator straight x squared plus 2 straight v squared straight x squared end fraction space space space or space space space straight v plus straight x dv over dx space equals space minus fraction numerator 2 straight v over denominator 1 plus 2 straight v squared end fraction therefore space space space space space space straight x dv over dx equals negative fraction numerator 2 straight v over denominator 1 plus 2 straight v squared end fraction minus straight v space space space space space or space space space space space straight x dv over dx equals space fraction numerator negative 2 straight v minus straight v minus 2 straight v cubed over denominator 1 plus 2 straight v squared end fraction therefore space space space space space straight x dv over dx space equals fraction numerator negative 3 straight v minus 2 straight v cubed over denominator 1 plus 2 straight v squared end fraction space space space space space space space rightwards double arrow space space space space space space fraction numerator 1 plus 2 straight v squared over denominator 3 straight v plus 2 straight v cubed end fraction dv space equals space minus 1 over straight x dx

Integrating,

1 third integral fraction numerator 3 plus 6 straight v squared over denominator 3 straight v plus 2 straight v cubed end fraction space equals space minus integral 1 over straight x dx

therefore space space space 1 third space log space open vertical bar 3 straight v space plus space 2 straight v cubed close vertical bar space equals space minus 3 space log space open vertical bar straight x close vertical bar space plus space 3 space log space open vertical bar straight c subscript 1 close vertical bar therefore space space space log space open vertical bar 3 straight v plus 2 straight v cubed close vertical bar space plus space log space open vertical bar straight x close vertical bar cubed space equals space log space left parenthesis straight c right parenthesis therefore space space space space open square brackets left parenthesis 3 space straight v plus space 2 straight v cubed right parenthesis space left parenthesis straight x right parenthesis close square brackets cubed space equals space straight c therefore space space space open parentheses 3 straight y over straight x plus 2 straight y cubed over straight x cubed close parentheses space left parenthesis straight x cubed right parenthesis space equals straight c or space space space 3 straight x squared straight y plus 2 straight y cubed space equals space straight c comma space which space is space required space solution. space

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