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Vector Algebra

Question
CBSEENMA12033069

Solve the following initial value problem:
(1 + x y) y dx + (1 – x y) x dy = 0, y (1) = 1.

Solution
The given differential equation is
                    (1 + x y) y dx + (1 – x y) x dy = 0
or            left parenthesis straight y space dx space plus space straight x space dy right parenthesis space plus space straight x space straight y squared space dx space minus space straight x squared straight y space dy space equals space 0
or            fraction numerator ydx plus xdy over denominator straight x squared straight y squared end fraction plus 1 over straight x dx minus 1 over straight y dy space equals space 0
therefore space space space space space integral fraction numerator straight y space dx space plus space straight x space dy over denominator straight x squared straight y squared end fraction plus integral 1 over straight x dx space minus space integral 1 over straight y dy space equals space 0               ...(1)
Let I = integral fraction numerator straight y space dx space plus space straight x space dy over denominator straight x squared straight y squared end fraction
Put x y  = t so that  x dy + y dx = dt
therefore space space space space straight I space equals space integral 1 over straight t squared dt space equals space integral straight t to the power of negative 2 end exponent dt space equals space fraction numerator straight t to the power of negative 1 end exponent over denominator negative 1 end fraction space equals space minus 1 over straight t space equals negative fraction numerator 1 over denominator straight x space straight y end fraction
therefore  from (1),
        negative 1 over xy plus logx minus logy space equals space straight c                             
Now     straight y left parenthesis 1 right parenthesis space equals space 1 space 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 1 space space space space when space straight x space equals space 1
therefore space space space minus 1 over 1 plus log space 1 space minus space log space 1 space equals space straight c space space space rightwards double arrow space space space space straight c space equals space minus 1
therefore space space space space from space left parenthesis 1 right parenthesis comma space solution space of space differential space equation space is space
space space space space space space space space space minus 1 over xy plus log space straight x space minus space log space straight y space equals space minus 1
therefore space space space log space straight x space equals space log space straight y space plus space fraction numerator 1 over denominator straight x space straight y end fraction minus 1
which is required solution.