Sponsor Area

Vector Algebra

Question
CBSEENMA12033006

Find a one-parameter family of solutions of the following differential equation, indicating carefully the interval in which the solutions are valid:
straight e to the power of straight y dx plus straight e to the power of straight x dy space equals space 0




Solution
The given differential equation is
        straight e to the power of straight x dx plus straight e to the power of straight x dy space equals space 0 space space space space space space space space space space space space space space space space or space space space space straight e to the power of straight x dy space equals space minus straight e to the power of straight x dx
therefore space space space space space space straight e to the power of straight y dy space equals space minus straight e to the power of straight x dx
Integration,  integral straight e to the power of negative straight y end exponent space dy space equals negative integral straight e to the power of negative straight x end exponent dx
therefore space space space space space fraction numerator straight e to the power of negative straight y end exponent over denominator negative 1 end fraction space equals negative fraction numerator straight e to the power of negative straight x end exponent over denominator negative 1 end fraction plus straight c subscript 1 space or space space straight e to the power of straight y plus straight e to the power of straight x space equals space straight c comma space space straight x space element of space straight R is the solution.