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Differential Equations

Question
CBSEENMA12033115

Show that the following differential equation is homogeneous and find a primitive of it. Derive the solution wherever possible:
left parenthesis 2 straight x squared straight y plus straight y cubed right parenthesis space dx plus space left parenthesis xy squared minus 3 straight x cubed right parenthesis space dy space equals space 0

Solution

The given differential equation is
   left parenthesis 2 straight x squared straight y plus straight y cubed right parenthesis space dx space plus space left parenthesis straight x space straight y squared minus space 3 space straight x cubed right parenthesis space dy space equals space 0
or                        dy over dx space equals space minus fraction numerator 2 straight x squared straight y plus straight y cubed over denominator xy squared minus 3 straight x cubed 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 straight v plus straight x dv over dx space equals space minus fraction numerator 2 straight x cubed straight v plus straight v cubed straight x cubed over denominator straight x cubed straight v squared minus 3 straight x cubed end fraction
therefore space space space space straight v plus straight x dv over dx space equals space fraction numerator 2 straight v plus straight v cubed over denominator 3 minus straight v squared end fraction
therefore space space space space space space straight x dv over dx space equals space fraction numerator 2 straight v cubed minus straight v over denominator 3 minus straight v squared end fraction
therefore space space space space fraction numerator 3 minus straight v squared over denominator 2 straight v cubed minus straight v end fraction dv space space equals space 1 over straight x dx 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 space space space space space space space space space space space space space space space space space space space space space.. left parenthesis 1 right parenthesis
therefore space space integral fraction numerator 3 minus straight v squared over denominator straight v left parenthesis 2 straight v squared minus 1 right parenthesis end fraction dv space equals space integral 1 over straight x dx
Put fraction numerator 3 minus straight v squared over denominator straight v left parenthesis 2 straight v squared minus 1 right parenthesis end fraction identical to space straight A over straight v plus fraction numerator Bv plus straight c over denominator 2 straight v squared minus 1 end fraction
therefore space space space space space space space 3 minus straight v squared space identical to space straight A space left parenthesis 2 straight v squared minus 1 right parenthesis space plus space Bv squared plus straight C space straight v                           ...(2)
Put v = 0 in (2)
therefore space space space space space space space 3 minus 0 space equals space straight A left parenthesis 0 minus 1 right parenthesis plus 0 plus 0 space space space space space space rightwards double arrow space space space 3 equals space space minus straight A space space space space space space rightwards double arrow space space space straight A space equals space minus 3
Equating   coefficients in (2) of
straight v squared right parenthesis            -1 = 2 A + B        rightwards double arrow space space minus 1 space equals space minus 6 plus straight B space space space space space space space space space space rightwards double arrow space space space straight B space equals space 5
straight v right parenthesis space space space space space space space space space space 0 space equals space straight c space space space space space space space space space space rightwards double arrow space space space space straight C space equals space 0
therefore space space fraction numerator 3 minus straight v squared over denominator straight v left parenthesis 2 straight v squared minus 1 right parenthesis end fraction identical to fraction numerator negative 3 over denominator straight v end fraction plus fraction numerator 5 straight v over denominator 2 straight v squared minus 1 end fraction
therefore space space space from space left parenthesis 1 right parenthesis space we space get comma
therefore space space space space minus 3 space integral 1 over straight v plus 5 over 4 integral fraction numerator 4 straight v over denominator 2 straight v squared minus 1 end fraction dv space equals space integral 1 over straight x dx
therefore space space space space minus 3 space log space open vertical bar straight v close vertical bar space plus space 5 over 4 log space open vertical bar 2 space straight v squared minus 1 close vertical bar space equals space log space open vertical bar straight x close vertical bar space plus space log space straight c subscript 1
space space space space space space space space space space minus 12 space log space open vertical bar straight v close vertical bar space plus space 5 space log space open vertical bar 2 straight v squared minus 1 close vertical bar space equals space 4 space log space open vertical bar straight x close vertical bar space space plus space 4 space log space straight c subscript 1
therefore space space space minus 12 space log space open vertical bar straight y over straight x close vertical bar space plus space 5 space log space open vertical bar fraction numerator 2 straight y squared over denominator straight x squared end fraction minus 1 close vertical bar space equals space 4 space log space open vertical bar straight x close vertical bar space plus space space 4 space log space straight c subscript 1
therefore space space space minus 12 space log space open vertical bar straight y close vertical bar space plus space 12 space log space open vertical bar straight x close vertical bar space plus space 5 space log space open vertical bar 2 straight y squared minus straight x squared close vertical bar space minus space 10 space log space open vertical bar straight x close vertical bar space minus space 4 space log space open vertical bar straight x close vertical bar space equals space 4 space log space straight c subscript 1
therefore space space space space minus 12 space log space open vertical bar straight y close vertical bar minus 2 space log space open vertical bar straight x close vertical bar space plus space 5 space log space open vertical bar 2 straight y squared minus straight x squared close vertical bar space equals space log space straight c
therefore space space space log space open vertical bar fraction numerator 2 straight y squared minus straight x squared over denominator straight y to the power of 12 straight x squared end fraction close vertical bar space equals space log space straight c
therefore space space space space space space space space fraction numerator 2 straight y squared minus straight x squared over denominator straight x squared space straight y to the power of 12 end fraction space equals space straight c
therefore space space space space space space space 2 space straight y squared minus straight x squared space equals space straight c space straight x squared space straight y to the power of 12 space is space the space required space solution. space

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