Differential Equations

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Question

For the given differential equation, find the particular solution satisfying the given condition:
$2 xy plus straight y squared minus 2 straight x squared straight y apostrophe space equals space 0 space space space space space space space space space space space space space straight y left parenthesis 1 right parenthesis space equals space 2$

Solution

The given differential equation is

2 xy plus straight y squared minus 2 straight x squared straight y apostrophe space equals space 0 space space space space space space space or space space space space space 2 straight x squared straight y apostrophe space equals space 2 xy plus straight y squared

therefore space space space space space space space space space dy over dx space equals space fraction numerator 2 xy plus straight y squared over denominator 2 straight x 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 space 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 straight v plus straight x dv over dx equals fraction numerator 2 straight v plus straight v squared over denominator 2 end fraction therefore space space space space space space space space straight x dv over dx equals fraction numerator 2 straight v plus 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 equals straight v squared over 2 therefore space space space space space 2 1 over straight v squared dv space equals space 1 over straight x dx

Integration,

2 space 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 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 minus 2 straight x over straight y space equals space log space open vertical bar straight x close vertical bar plus straight c

Now,

straight y left parenthesis 1 right parenthesis space equals space 2 space space space space space space space space space space space space space space rightwards double arrow space space space space straight y space equals space 2 space space space when space straight x space equals space 1

therefore space space space space minus 2. space 1 half space equals space log space 1 plus straight c space space space space space space space space space space rightwards double arrow space space space space minus 1 space equals space 0 space plus straight c space space space space rightwards double arrow space space space space straight c space equals space minus 1 therefore space space solution space is 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 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.