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Vector Algebra

Question
CBSEENMA12032893

For problem given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation:straight y equals straight e to the power of straight x left parenthesis straight a space cosx space plus space straight b space sin space straight x right parenthesis space colon thin space fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 dy over dx plus 2 straight y space equals space 0
         

Solution

The given differential equation is
y = ex (a cos x + b sin x)
⇒    e–x y = a cos x + b sin x    ...(1)
Differentiating (1) twice w.r.t. .x. we get
                  straight e to the power of negative straight x end exponent dy over dx plus straight y space straight e to the power of negative straight x end exponent left parenthesis negative 1 right parenthesis space equals space minus straight a space sinx space plus space straight b space cosx
and        straight e to the power of negative straight x end exponent fraction numerator straight d squared straight y over denominator dx squared end fraction plus dy over dx straight e to the power of negative straight x end exponent left parenthesis negative 1 right parenthesis minus open curly brackets negative ye to the power of negative straight x end exponent plus straight e to the power of negative straight x end exponent dy over dx close curly brackets space equals space minus straight a space cosx space minus space straight b space sinx
or            straight e to the power of negative straight x end exponent fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 straight e to the power of negative straight x end exponent dy over dx plus straight y space straight e to the power of negative straight x end exponent space equals space minus straight y space straight e to the power of negative straight x end exponent                     [because space of space left parenthesis 1 right parenthesis]
or         straight e to the power of negative straight x end exponent fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 straight e to the power of negative straight x end exponent dy over dx plus 2 space straight y space straight e to the power of negative straight x end exponent space equals space 0 space space or space space straight e to the power of negative straight x end exponent open curly brackets fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 dy over dx plus 2 straight y close curly brackets space equals space 0
or            fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 dy over dx plus 2 straight y space equals space 0
Hence the result.