Differential Equations

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Question

For the given differential equation, find the particular solution satisfying the given condition:
$2 xy plus straight y squared minus 2 straight x squared dy over dx equals 0 semicolon space space space straight y space equals 2 space space when space straight x space equals space 1.$

Solution

The given differential equation is

2 xy plus straight y minus 2 straight x squared dy over dx space equals space 0 space space space space space space space space space or space space space space space space 2 straight x squared dy over dx space equals space 2 xy plus straight y squared

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

Put

straight y equals space straight v space straight x space space so space that space dy over dx space equals space straight v plus straight x dv over dx

therefore

straight v plus straight x dv over dx equals fraction numerator 2 straight x squared straight v plus straight v squared straight x squared over denominator 2 straight x squared end fraction

therefore space space space space space space straight v plus straight x dv over dx space equals fraction numerator 2 straight v plus straight v squared over denominator 2 end fraction therefore space space space space space space space space space space straight x dv over dx space equals space fraction numerator 2 straight v plus straight v squared over denominator 2 end fraction minus straight v space space space space or space space space straight x dv over dx equals straight v squared over 2 therefore space space space space space space 2 space 1 over straight v squared dv space equals space 1 over straight x dx

Integrating,

2 integral straight v to the power of negative 2 end exponent space dv space equals space integral 1 over straight x dx

therefore space space space space space 2 fraction numerator straight v to the power of negative 1 end exponent over denominator negative 1 end fraction space equals space log space open vertical bar straight x close vertical bar plus space straight c space space space space space or space space space space fraction numerator negative 2 over denominator straight v end fraction space equals space log space open vertical bar straight x close vertical bar plus straight c therefore space space space space space minus 2 straight x over straight y space equals log space open vertical bar straight x close vertical bar space plus space straight c space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis

Now v = 2 when x = 1

therefore space space space space space space space space space space space space space space minus 2. space 1 half minus log space 1 space plus space straight c space space space space space space space rightwards double arrow space space space minus 1 space equals space 0 plus straight c space space space space rightwards double arrow space space space straight c space equals space minus 1 therefore space space space space space from space left parenthesis 1 right parenthesis comma space space space minus fraction numerator 2 straight x over denominator straight y end fraction space equals space log space open vertical bar straight x close vertical bar minus 1 space space space space space or space space space space straight y space equals space fraction numerator 2 straight x over denominator 1 minus log space open vertical bar straight x close vertical bar end fraction is space the space required space solution. space