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Differential Equations

Question
CBSEENMA12033112

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
(x2 + y2) y' = 8 x2 - 3 x y + 2 y2

Solution
The given differential equation is
                   left parenthesis straight x squared plus straight y squared right parenthesis space dy over dx space equals space 8 straight x squared minus 3 xy plus 2 straight y squared
therefore                           dy over dx space equals space fraction numerator 8 straight x squared minus 3 xy plus 2 straight y squared over denominator straight x squared plus straight y squared end fraction
Put straight y equals vx space so space that space dy over dx space equals space straight v plus straight x dv over dx
therefore space space space space space space space straight v plus straight x dv over dx space equals space fraction numerator 8 straight x squared minus 3 straight x cubed straight v plus 2 straight x squared straight v squared over denominator straight x squared plus straight v squared straight x squared end fraction
therefore space space space space space straight v plus straight x dv over dx space equals space fraction numerator 8 minus 3 straight v plus 2 straight v squared over denominator 1 plus straight v squared end fraction
therefore space space space space space space space space space space space straight x dv over dx space equals space fraction numerator 8 minus 3 straight v plus 2 straight v squared over denominator 1 plus straight v squared end fraction minus straight v
therefore space space space space space space space space straight x dv over dx space equals space fraction numerator 8 minus 3 straight v plus 2 straight v squared minus straight v minus straight v cubed over denominator 1 plus straight v squared end fraction
therefore space space space space space straight x dv over dx space equals space fraction numerator 8 minus 4 straight v plus 2 straight v squared minus straight v cubed over denominator 1 plus straight v squared end fraction
Separating the variables, we get,
                fraction numerator 1 plus straight v squared over denominator 8 minus 4 straight v plus 2 straight v squared minus straight v cubed end fraction dv space equals space 1 over straight x dx
Integrating,  integral fraction numerator 1 plus straight v squared over denominator 8 minus 4 straight v plus 2 straight v squared minus straight v cubed end fraction dv space equals space integral 1 over straight x dx
therefore space space space space integral fraction numerator 1 plus straight v squared over denominator left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 4 plus straight v squared right parenthesis end fraction dv space equals space integral 1 over straight x dx
Put fraction numerator 1 plus straight v squared over denominator left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 4 plus straight v squared right parenthesis end fraction space equals space fraction numerator straight A over denominator 2 minus straight v end fraction plus fraction numerator Bv plus straight c over denominator 4 plus straight v squared end fraction
therefore          1 plus straight v squared space space identical to space space straight A space left parenthesis 4 plus straight v squared right parenthesis space plus space straight B space straight v thin space left parenthesis 2 minus straight v right parenthesis space plus space straight C left parenthesis 2 minus straight v right parenthesis                ...(1)
Put             2 minus straight v space equals space 0 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 or space space space space space space straight v space equals space 2 space in space left parenthesis 1 right parenthesis
therefore space space space space space space space space space space space 1 plus 4 space equals space straight A left parenthesis 4 plus 4 right parenthesis plus 0 plus 0 space space space space space space space space space space space space space space space rightwards double arrow space space 5 space equals space 8 straight A space space space space space space space rightwards double arrow space space straight A space equals space 5 over 8
(1) can be written as
                                1 plus straight v squared identical to space straight A left parenthesis 4 plus straight v squared right parenthesis space plus space straight B left parenthesis 2 straight v minus straight v squared right parenthesis space plus straight C left parenthesis 2 minus straight v right parenthesis        ...(2)
Equating coefficients in (2) of
 V2)              1 =A - B                  rightwards double arrow space space space space 1 space equals space 5 over 8 minus straight B space space space space space space space space rightwards double arrow space space space space straight B space equals space minus 3 over 8
v)                   0 = 2 B - C             rightwards double arrow space 0 space equals space 3 over 4 minus straight C space space space space space space rightwards double arrow space space space space straight C space equals space minus 3 over 4
therefore space space space space fraction numerator 1 plus straight v squared over denominator left parenthesis 2 minus straight v right parenthesis thin space left parenthesis 4 plus straight v squared right parenthesis end fraction space identical to space space fraction numerator 5 over denominator 8 space left parenthesis 2 minus straight v right parenthesis end fraction plus fraction numerator negative begin display style 3 over 8 end style straight v minus begin display style 3 over 4 end style over denominator 4 plus straight v end fraction
therefore space space from space left parenthesis 1 right parenthesis comma space space integral open square brackets fraction numerator 5 over denominator 8 left parenthesis 2 minus straight v right parenthesis end fraction plus fraction numerator negative begin display style 3 over 8 end style straight v minus begin display style 3 over 4 end style over denominator straight v squared plus 4 end fraction close square brackets space dv space equals space integral 1 over straight x dx
therefore space space space 5 over 8 integral fraction numerator 1 over denominator 2 minus straight v end fraction dv space minus 3 over 16 integral fraction numerator 2 straight v over denominator straight v squared plus 4 end fraction dv minus 3 over 4 integral fraction numerator 1 over denominator straight v squared plus left parenthesis 2 right parenthesis squared end fraction dv space equals space integral 1 over straight x dx
therefore space space space 5 over 8 fraction numerator log space open vertical bar 2 minus straight v close vertical bar over denominator negative 1 end fraction minus 3 over 16 log space left parenthesis straight v squared plus 4 right parenthesis minus space 3 over 4 space 1 half space tan to the power of negative 1 end exponent straight v over 2 space equals space log space open vertical bar straight x close vertical bar space space plus straight c
therefore space space space minus 5 over 8 log space open vertical bar 2 minus straight y over straight x close vertical bar minus 3 over 16 log space open parentheses straight y squared over straight x squared plus 4 close parentheses space minus space 3 over 8 tan to the power of negative 1 end exponent open parentheses fraction numerator straight y over denominator 2 straight x end fraction close parentheses space equals space log space open vertical bar straight x close vertical bar plus straight c
which is required primitive.


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