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Vector Algebra

Question
CBSEENMA12033057

Solve:
straight x squared left parenthesis straight y minus 1 right parenthesis space dx space plus space straight y squared space left parenthesis straight x minus 1 right parenthesis space dy space equals space 0

 

Solution

The given differential equation is
             straight x squared left parenthesis straight y minus 1 right parenthesis space dx space plus space straight y squared left parenthesis straight x minus 1 right parenthesis space dy space equals space 0 space space or space space straight y squared left parenthesis straight x minus 1 right parenthesis space dy space equals space minus straight x squared left parenthesis straight y minus 1 right parenthesis space dx
or      fraction numerator straight y squared over denominator straight y minus 1 end fraction dy space equals space minus fraction numerator straight x squared over denominator straight x minus 1 end fraction dx
Integrating,   integral fraction numerator straight y squared over denominator straight y minus 1 end fraction dy space equals space minus integral fraction numerator straight x squared over denominator straight x minus 1 end fraction dx
therefore space space space space space integral open parentheses straight y plus 1 plus fraction numerator 1 over denominator straight y minus 1 end fraction close parentheses dy space equals space minus integral open parentheses straight x plus 1 plus fraction numerator 1 over denominator straight x minus 1 end fraction close parentheses dx
therefore space space straight y squared over 2 plus straight y plus log space open vertical bar straight y minus 1 close vertical bar space equals space minus open square brackets straight x squared over 2 plus straight x plus log space open vertical bar straight x plus 1 close vertical bar close square brackets plus straight c
is the required solution.