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Vector Algebra

Question
CBSEENMA12033063

Solve
dy over dx space equals space fraction numerator xe to the power of straight x logx plus straight e to the power of straight x over denominator straight x space cosy end fraction

Solution

The given differential equation is dy over dx equals fraction numerator xe to the power of straight x logx plus straight e to the power of straight x over denominator straight x space cosy end fraction
Separating the variables, we get,
            cosy space dy space equals space fraction numerator xe to the power of straight x logx plus straight e to the power of straight x over denominator straight x end fraction dx space space or space space cosy space dy space equals space straight e to the power of straight x open parentheses fraction numerator straight x space logx space plus space 1 over denominator straight x end fraction close parentheses dx
or     cosy space dy space equals space straight e to the power of straight x open parentheses logx plus 1 over straight x close parentheses dx space space space rightwards double arrow space space integral space cosy space dy space equals space integral straight e to the power of straight x open parentheses logx plus 1 over straight x close parentheses dx
therefore space space space sin space straight y space equals space straight e to the power of straight x space logx space plus space straight c space space space space space space space space space open square brackets because space space integral straight e to the power of straight x open curly brackets straight f left parenthesis straight x right parenthesis plus straight f apostrophe left parenthesis straight x right parenthesis close curly brackets dx space equals space straight e to the power of straight x space straight f left parenthesis straight x right parenthesis space plus space straight c close square brackets
which is required solution.