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Vector Algebra

Question
CBSEENMA12032962

Obtain the differential equation from  the equation y = ex (a cos 2x + b sin 2x). where a and b are arbitrary constants.    

Solution

The given equation is
                      y = ex (a cos 2x + b sin 2x)     ... (1)
Differentiating both sides w.r.t x, we get,
                     dy over dx space equals space straight e to the power of straight x left parenthesis straight a space cos space 2 straight x space plus space straight b space sin space 2 straight x right parenthesis space plus space straight e to the power of straight x left parenthesis negative 2 straight a space sin space 2 straight x space plus space 2 straight b space cos space 2 straight x right parenthesis    
therefore space space space space dy over dx space equals space straight y plus straight e to the power of straight x left parenthesis negative 2 straight a space sin space 2 straight x space plus space 2 straight b space cos space 2 straight x right parenthesis    ...(2)
                                                                                  open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
Again differentiating w.r.t x,
       space space space space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals dy over dx plus straight e to the power of straight x left parenthesis negative 2 straight a space sin space 2 straight x space plus space 2 straight b space cos space 2 straight x right parenthesis space plus space straight e to the power of straight x left parenthesis negative 4 straight a space cos space 2 straight x space minus space 4 straight b space sin space 2 straight x right parenthesis
or          fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space dy over dx plus open parentheses dy over dx minus straight y close parentheses minus 4 straight e to the power of straight x left parenthesis straight a space cos space 2 straight x space plus space straight b space sin space 2 straight x right parenthesis space space space space space space open square brackets because space of space left parenthesis 2 right parenthesis close square brackets
or          fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space dy over dx plus dy over dx minus straight y minus 4 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 open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
or          fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 dy over dx plus 5 straight y space equals space 0
which is required differential equation.