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Vector Algebra

Question
CBSEENMA12032964

Find the differential equation of the family of curves y = a sin (bx + c). a, b. c being arbitrary constants.

Solution
The given equation is
                       y = a sin (bx + c)                             ...(1)
Differentiating w.r.t.x, successively three times,
                     dy over dx space equals straight b space straight a space cos left parenthesis bx plus straight c right parenthesis                     ...(2)
            fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space minus straight b squared straight a space sin left parenthesis bx plus straight c right parenthesis                       ...(3)
         fraction numerator straight d cubed straight y over denominator dx cubed end fraction space equals space minus straight b cubed straight a space cos left parenthesis bx plus straight c right parenthesis                         ...(4)
From (1) and (2), (3) and (4), we get,
             straight y fraction numerator straight d cubed straight y over denominator dx cubed end fraction equals negative straight b cubed straight a squared space sin space left parenthesis bx plus straight c right parenthesis space cos space left parenthesis bx plus straight c right parenthesis
                            equals space left square bracket straight b space straight a space cos space left parenthesis bx plus straight c right parenthesis right square bracket space left square bracket straight b squared straight a space sin left parenthesis bx plus straight c right parenthesis right square bracket
or           straight y fraction numerator straight d cubed straight y over denominator dx cubed end fraction space equals space open parentheses dy over dx close parentheses space fraction numerator straight d squared straight y over denominator dx squared end fraction is the required differential equation.