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Differential Equations

Question

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Solve the following differential equation:
$straight x squared dy over dx space equals space 2 xy plus straight y squared$

Solution

The given differential equation is

straight x squared dy over dx space equals space 2 xy plus straight y squared

or

dy over dx space equals space fraction numerator 2 xy plus straight y squared over denominator straight x squared end fraction

...(1)
Put

straight y space equals space straight v space straight x space space space space so space that space space space dy over dx equals straight v plus 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 space equals space fraction numerator 2 straight x. space vx plus straight v squared straight x squared over denominator straight x squared end fraction

or

straight v plus straight x space dv over dx space equals space 2 space straight v space plus space straight v squared

or

straight x dv over dx space equals straight v squared plus straight v

Separating the variables, we get,

fraction numerator 1 over denominator straight v squared plus straight v end fraction dv space equals space 1 over straight x dx

Integrating,

integral fraction numerator 1 over denominator straight v squared plus straight v end fraction dv space equals space integral 1 over straight x dx

therefore space space space integral fraction numerator 1 over denominator straight v left parenthesis straight v plus 1 right parenthesis end fraction dv space equals space integral 1 over straight x dx

space therefore space space space space integral open square brackets fraction numerator 1 over denominator straight v left parenthesis 0 plus 1 right parenthesis end fraction plus fraction numerator 1 over denominator left parenthesis negative 1 right parenthesis thin space left parenthesis straight v plus 1 right parenthesis end fraction close square brackets space dv space equals space integral space 1 over straight x dx therefore space space space space integral open parentheses 1 over straight v minus fraction numerator 1 over denominator straight v plus 1 end fraction close parentheses space dv space equals space integral 1 over straight x dx therefore space space space log space open vertical bar straight v close vertical bar space minus space log space open vertical bar straight v plus 1 close vertical bar space equals space log space open vertical bar straight x close vertical bar space plus space log space straight c subscript 1 therefore space space log space open vertical bar fraction numerator straight v over denominator straight v plus 1 end fraction close vertical bar space equals space log space left parenthesis straight c subscript 1 space open vertical bar straight x close vertical bar right parenthesis therefore space space open vertical bar fraction numerator straight v over denominator straight v plus 1 end fraction close vertical bar space equals space straight c subscript 1 open vertical bar straight x close vertical bar therefore space space space space space fraction numerator begin display style straight y over straight x end style over denominator begin display style straight y over straight x plus 1 end style end fraction equals space space cx therefore space space space space fraction numerator straight y over denominator straight x plus straight y end fraction space equals space straight c space straight x comma space which space is space required space solution. space

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