Question

For the given differential equation, find the particular solution satisfying the given condition:
$2 straight x squared straight y apostrophe space minus space 2 xy plus straight y squared space equals space 0 comma space space space space straight y left parenthesis straight e right parenthesis space equals space straight e$

Solution

The given differential equation is
$2 straight x squared straight y apostrophe space minus space 2 xy plus straight y squared space equals space 0$
or $2 straight x squared dy over dx space equals space 2 xy minus straight y squared space space space space space or space space space space dy over dx space equals space fraction numerator 2 xy minus straight y squared over denominator 2 straight x squared end fraction$
Put y = v x so that $dy over dx space equals straight v plus straight x dv over dx$
$therefore space space space space space straight v plus straight x dv over dx space equals space fraction numerator 2 straight x squared straight v minus straight v squared straight x squared over denominator 2 straight x squared end fraction$
$therefore space space space space straight v plus straight x dv over dx equals fraction numerator 2 straight v minus straight v squared over denominator 2 end fraction therefore space space space space space space straight x dv over dx space equals space fraction numerator 2 straight v minus straight v squared over denominator 2 end fraction minus straight v space space space space space or space space space straight x dv over dx space equals negative straight v squared over 2 therefore space space space space space space minus 2 over straight v squared dv space equals space 1 over straight x dx$
Integrating $2 integral straight v to the power of negative 2 end exponent dv space equals space integral 1 over straight x dx$
$therefore space space space minus 2 fraction numerator straight v to the power of negative 1 end exponent over denominator negative 1 end fraction space equals space log space open vertical bar straight x close vertical bar space plus space straight c space space space space space or space space space space fraction numerator negative 2 over denominator straight v end fraction equals log space open vertical bar straight x close vertical bar plus straight c therefore space space space space space space fraction numerator 2 straight x over denominator straight y end fraction space equals space log space open vertical bar straight x close vertical bar space plus space straight c$
Now $straight y left parenthesis straight e right parenthesis space equals straight e$ $rightwards double arrow space space space straight y space equals space straight e space space space when space straight x space equals space straight e$
$therefore space space space 2 straight e over straight e equals log space straight e plus straight c space space space space space space space space space space rightwards double arrow space space space space 2 space equals space 1 plus straight c space space space space space space space rightwards double arrow space space space straight c space equals space 1$
$therefore space space space solution space is space space fraction numerator 2 straight x over denominator straight y end fraction space equals space log space open vertical bar straight x close vertical bar plus 1.$

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