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Differential Equations

Question
CBSEENMA12033130
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Find a one parameter family of solutions of each of the following differential equation:
y2 dx + (x2 – x y + y2) dy = 0

Solution
The given differential equation is
              y2 dx + (x2 – x y + y2) dy = 0      or      (x2 – x y + y2) dy = -  y2 dx
therefore space space space space space space space space space space space space space space space space space space space space dy over dx equals negative fraction numerator straight y squared over denominator straight x squared minus xy plus straight y squared end fraction
Put y = vx 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 negative fraction numerator straight v squared straight x squared over denominator straight x squared minus vx squared plus straight v squared straight x squared end fraction therefore space space space space space straight v plus straight x dv over dx equals negative fraction numerator straight v squared over denominator 1 minus straight v plus straight v squared end fraction therefore space space space space space space straight x dv over dx space equals space minus fraction numerator straight v squared over denominator 1 minus straight v plus straight v squared end fraction minus straight v therefore space space space space space straight x dv over dx equals fraction numerator negative straight v squared minus straight v plus straight v squared minus straight v cubed over denominator 1 minus straight v plus straight v squared end fraction therefore space space space space space space straight x dv over dx equals fraction numerator negative straight v minus straight v cubed over denominator 1 minus straight v plus straight v squared end fraction therefore space space fraction numerator 1 minus straight v plus straight v squared over denominator straight v plus straight v cubed end fraction dv space equals space minus 1 over straight x dx therefore space space space space space integral fraction numerator 1 minus straight v plus straight v squared over denominator straight v left parenthesis 1 plus straight v squared right parenthesis end fraction space identical to space straight A over straight v plus fraction numerator Bv plus straight c over denominator 1 plus straight v squared end fraction
     Multiplying both sides by v (1 + v2), we get.
                        1 minus straight v plus straight v squared space equals space straight A thin space left parenthesis 1 plus straight v squared right parenthesis space plus space Bv squared plus Cv                  ...(1)
Put                       v = 0 in (1)
therefore space space space space space space 1 minus 0 plus 0 space equals space straight A space left parenthesis 1 plus 0 right parenthesis space plus space 0 space plus space 0. space space space space space space rightwards double arrow space space space straight A space equals space 1
Equating coefficients in (1) of
straight v squared right parenthesis               1 = A + B                 rightwards double arrow 1 space equals 1 space plus space straight B space space space space space space space space space space space space space space space space rightwards double arrow space space space straight B space equals space 0
v)                 -1 = C                       rightwards double arrow space straight C space equals space minus 1
therefore space space space space space fraction numerator 1 minus straight v plus straight v squared over denominator straight v left parenthesis 1 plus straight v squared right parenthesis end fraction identical to 1 over straight v plus fraction numerator negative 1 over denominator 1 plus straight v squared end fraction
therefore space space space space space from space left parenthesis 1 right parenthesis comma space we space get
             integral open parentheses 1 over straight v minus fraction numerator 1 over denominator 1 plus straight v squared end fraction close parentheses space dv space equals space minus integral 1 over straight x dx
therefore space space log space open vertical bar straight v close vertical bar space minus space tan to the power of negative 1 end exponent straight v space equals space space minus space log space open vertical bar straight x close vertical bar space plus space straight c therefore space space space log space open vertical bar straight y over straight x close vertical bar space minus space tan to the power of negative 1 end exponent straight y over straight x space equals space minus log space open vertical bar straight x close vertical bar space plus space log space straight A therefore space space space log space open vertical bar straight y close vertical bar space minus space log space open vertical bar straight x close vertical bar space minus space tan to the power of negative 1 end exponent straight y over straight x space equals space minus log space open vertical bar straight x close vertical bar plus space space log space straight A therefore space space space log space open vertical bar straight y close vertical bar minus log space straight A space equals space space tan to the power of negative 1 end exponent straight y over straight x therefore space space space log space open vertical bar fraction numerator open vertical bar straight y close vertical bar over denominator straight A end fraction close vertical bar space equals tan to the power of negative 1 end exponent straight y over straight x space space space space rightwards double arrow space space space space fraction numerator open vertical bar straight y close vertical bar over denominator straight A end fraction space equals space straight e to the power of tan to the power of negative 1 end exponent straight y over straight x end exponent therefore space space space space space space space space space space space space space space space space space space open vertical bar straight y close vertical bar space equals space Ae to the power of tan to the power of negative 1 end exponent straight y over straight x end exponent therefore space space space space space space space space space space space space space space space space space space straight y squared space equals space space straight c space straight e space to the power of 2 space tan to the power of negative 1 end exponent straight y over straight x end exponent
is the required solution. 

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