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Vector Algebra

Question
CBSEENMA12033078

For the differential equation xy dy over dx space equals space left parenthesis straight x plus 2 right parenthesis thin space left parenthesis straight y plus 2 right parenthesis comma find the solution curve passing through the point (1, -1).

Solution
The given differential equation is
        straight x space straight y dy over dx space equals space left parenthesis straight x plus 2 right parenthesis thin space left parenthesis straight y plus 2 right parenthesis
Separating the variables, we get,
                    fraction numerator straight y over denominator straight y plus 2 end fraction dy space equals space fraction numerator straight x plus 2 over denominator straight x end fraction dx
Integrating,   integral fraction numerator straight y over denominator straight y plus 2 end fraction dy space equals space integral fraction numerator straight x plus 2 over denominator straight x end fraction dx
therefore space space space space integral fraction numerator left parenthesis straight y plus 2 right parenthesis space minus space 2 over denominator straight y plus 2 end fraction dy space equals space integral open parentheses straight x over straight x plus 2 over straight x close parentheses dx
therefore space space space space integral open parentheses 1 minus fraction numerator 2 over denominator straight y plus 2 end fraction close parentheses space dy space equals space integral open parentheses 1 plus 2 over straight x close parentheses dx
therefore space space space straight y minus 2 space open vertical bar straight y plus 2 close vertical bar space equals space straight x plus 2 space log open vertical bar straight x close vertical bar space plus space straight c                       ...(1)
Since the curve passes through (1, -1)
therefore space space space space space minus 1 minus 2 space log space open vertical bar negative 1 plus 2 close vertical bar space equals space 1 plus 2 space log space open vertical bar 1 close vertical bar space plus space straight c
therefore space space space space space space minus 1 minus 2 space log space open vertical bar 1 close vertical bar space equals space 1 plus 2 space log space open vertical bar 1 close vertical bar plus straight c
therefore space space space minus 1 minus 2 space left parenthesis 0 right parenthesis space equals space 1 space plus space 2 left parenthesis 0 right parenthesis space plus space straight c space space space rightwards double arrow space space straight c space equals space minus 2
therefore space space space from space left parenthesis 1 right parenthesis comma space space space straight y minus 2 space log space open vertical bar straight y plus 2 close vertical bar space equals space straight x plus 2 space log space open vertical bar straight x close vertical bar space minus space 2
or        straight y minus straight x plus 2 space equals space 2 space log space open vertical bar straight x space left parenthesis straight y plus 2 right parenthesis close vertical bar
or           straight y minus straight x plus 2 space equals space log space open vertical bar straight x space left parenthesis straight y plus 2 right parenthesis close vertical bar squared
or            straight y minus straight x plus 2 space equals space log space open square brackets straight x squared space left parenthesis straight y plus 2 right parenthesis squared close square brackets
is the required solution.