Sponsor Area

Vector Algebra

Question
CBSEENMA12032986

Which of the following differential equations has y = c1ex + c2e–x as the general solution?
  • fraction numerator straight d squared straight y over denominator dx squared end fraction plus straight y space equals space 0
  • fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight y space equals space 0
  • fraction numerator straight d squared straight y over denominator dx squared end fraction plus 1 space equals space 0
  • fraction numerator straight d squared straight y over denominator dx squared end fraction minus 1 space equals space 0

Solution

B.

fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight y space equals space 0

The given equation is
                        straight y space equals straight c subscript 1 straight e to the power of straight x plus straight c subscript 2 straight e to the power of negative straight x end exponent                                ...(1)
therefore space space space space space dy over dx space equals space straight c subscript 1 straight e to the power of straight x minus straight c subscript 2 straight e 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 space equals space straight c subscript 1 straight e to the power of straight x plus straight c subscript 2 straight e to the power of negative straight x end exponent space space space rightwards double arrow space space space fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space straight y 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
rightwards double arrow space space space space space space fraction numerator straight d squared straight y over denominator dx squared end fraction minus straight y space equals space 0 comma space space which space is space the space required space differential space equation.
therefore   (B) is the correct answer.