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Vector Algebra

Question
CBSEENMA12032951
Wired Faculty App

Form the differential equation of the family of curves
straight y equals Ae to the power of straight x plus Be to the power of negative straight x end exponent
where A and B are constants.

Solution

The given equation is
                           straight y equals Ae to the power of straight x plus Be to the power of negative straight x end exponent                          ...(1)
therefore space space space dy over dx space equals space Ae to the power of straight x minus Be to the power of negative straight x end exponent 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 rightwards double arrow space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space Ae to the power of straight x plus Be to the power of negative straight x end exponent
rightwards double arrow space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction equals space straight y                                                       open square brackets because space of space left parenthesis 1 right parenthesis close square brackets
which is required differential equation. 

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