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Continuity And Differentiability

Question

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If =3 cos(log x)+4 sin(log x), show that x²y₂+x y₁+y=0.

Solution

straight y equals 3 space cos left parenthesis log space straight x right parenthesis plus 4 space sin left parenthesis right parenthesis log space straight x space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis therefore space space space space space space space space straight y subscript 1 equals negative fraction numerator 3 space sin left parenthesis log space straight x right parenthesis over denominator straight x end fraction plus fraction numerator 4 space cos left parenthesis log space straight x right parenthesis over denominator straight x end fraction therefore space space space space space straight x space straight y subscript 1 equals negative 3 space sin left parenthesis log space straight x right parenthesis plus 4 space cos left parenthesis log space straight x right parenthesis Again space differentiating space straight w. straight r. straight t. straight x comma space we space get comma space space space space space space space space space space space space space space straight x space straight y subscript 2 plus straight y subscript 1.1 equals negative fraction numerator 3 space sin left parenthesis log space straight x right parenthesis over denominator straight x end fraction minus fraction numerator 4 space cos left parenthesis log space straight x right parenthesis over denominator straight x end fraction therefore space space space space space space space space space straight x squared straight y subscript 1 plus xy subscript 1 equals negative left square bracket 3 space cos left parenthesis log space straight x right parenthesis plus 4 space sin left parenthesis log space straight x right parenthesis right square bracket therefore space space space space space space space space space straight x squared straight y subscript 1 plus xy subscript 1 equals negative straight y space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket therefore space space space straight x squared straight y subscript 2 plus xy subscript 1 plus straight y equals 0

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