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Differential Equations

Question
CBSEENMA12033110

Solve the following differential equation:
(y2 – x2) dy = 3 x y dx

Solution
The given differential equation is
                     open parentheses straight y squared minus straight x squared close parentheses space dy space equals space 3 xy space dx
or            dy over dx space equals space fraction numerator 3 space straight x space straight y over denominator straight y squared minus straight x squared end fraction                 ...(1)
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 space space left parenthesis 1 right parenthesis space becomes comma space space straight v plus straight x dv over dx space equals space fraction numerator 3 straight x. space straight v space straight x over denominator straight v squared straight x squared minus straight x squared end fraction
or                  straight v plus straight x dv over dx space equals space fraction numerator 3 straight v over denominator straight v squared minus 1 end fraction
therefore space space space space space space straight x space dv over dx space equals space fraction numerator 3 straight v over denominator straight v squared minus 1 end fraction minus straight v space space space space or space space space space straight x space dv over dx space equals space fraction numerator 4 straight v space minus straight v cubed over denominator straight v squared minus 1 end fraction
Separating the variables, we get,
                                  fraction numerator straight v squared minus 1 over denominator 4 space straight v minus straight v cubed end fraction dv space equals space 1 over straight x dx space space space or space space space fraction numerator straight v squared minus 1 over denominator straight v left parenthesis 4 minus straight v squared right parenthesis end fraction space dv space equals space 1 over straight x dx
Integrating,        integral fraction numerator straight v squared minus 1 over denominator straight v space left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 2 plus straight v right parenthesis end fraction dv space equals space integral 1 over straight x dx
therefore space space space space integral open square brackets fraction numerator 0 minus 1 over denominator straight v space left parenthesis 2 minus 0 right parenthesis thin space left parenthesis 2 plus 0 right parenthesis end fraction minus fraction numerator 4 minus 1 over denominator left parenthesis 2 right parenthesis thin space left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 2 plus 2 right parenthesis end fraction plus fraction numerator 4 minus 1 over denominator left parenthesis negative 2 right parenthesis thin space left parenthesis 2 plus 2 right parenthesis thin space left parenthesis 2 plus straight v right parenthesis end fraction close square brackets dv space equals space integral 1 over straight x dx
therefore space space integral open square brackets negative fraction numerator 1 over denominator 4 straight v end fraction plus fraction numerator 3 over denominator 8 left parenthesis 2 minus straight v right parenthesis end fraction minus fraction numerator 3 over denominator 8 left parenthesis 2 plus straight v right parenthesis end fraction close square brackets space dv space equals space integral 1 over straight x dx
therefore space space space space minus space 1 fourth open vertical bar straight v close vertical bar plus space 3 over 8 fraction numerator log space open vertical bar 2 minus straight v close vertical bar over denominator negative 1 end fraction space minus space 3 over 8 space log space open vertical bar 2 plus straight v close vertical bar space equals space log space straight x space plus straight c
therefore space space space space space minus 1 fourth space log space open vertical bar straight y over straight x close vertical bar space minus space 3 over 8 space log space open vertical bar 2 minus straight y over straight x close vertical bar space minus space 3 over 8 log space open vertical bar 2 plus straight y over straight x close vertical bar space equals space logx plus straight c
which is required solution. 

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