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Vector Algebra

Question
CBSEENMA12032958

Form a differential equation from the equation y = ae 2x + be– 3x , a , b being constants. 

Solution

The given equation is
                        y = ae 2x + be– 3x                            ...(1)
Differentiating both sides w.r.t.x, we get
                        dy over dx space equals space 2 ae to the power of 2 straight x end exponent minus 3 be to the power of negative 3 straight x end exponent               ...(2)
Again differentiation both sides w.r.t.x, we get.
          fraction numerator straight d squared straight y over denominator dx squared end fraction space equals space 4 ae to the power of 2 straight x end exponent plus 9 be to the power of negative 3 straight x end exponent                            ...(3)
Multiplying (2) by 2 and subtracting from (3), we get,
            fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 dy over dx space equals space 15 space be to the power of negative 3 straight x end exponent
therefore space space space space space be to the power of negative 3 straight x end exponent space equals 1 over 15 open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 dy over dx close parentheses                     ...(4)
Multiplying (2) by 3 and adding it to (3), we get,
                        fraction numerator straight d squared straight y over denominator dx squared end fraction plus 3 dy over dx space equals space 10 space straight a space straight e to the power of 2 straight x end exponent
therefore space space space space space space ae to the power of 2 straight x end exponent space equals space 1 over 10 open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction plus 3 dy over dx close parentheses                       ...(5)
From (1), (4) and (5), we get,
                             straight y equals space 1 over 10 open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction plus 3 dy over dx close parentheses plus 1 over 15 open parentheses fraction numerator straight d squared straight y over denominator dx squared end fraction minus 2 dy over dx close parentheses
or                       30 space straight y space equals space 3 space fraction numerator straight d squared straight y over denominator dx squared end fraction plus 9 dy over dx plus 2 fraction numerator straight d squared straight y over denominator dx squared end fraction minus 4 dy over dx
or             5 fraction numerator straight d squared straight y over denominator dx squared end fraction plus 5 dy over dx space equals space 30 space straight y space space space space space or space space space fraction numerator straight d squared straight y over denominator dx squared end fraction plus dy over dx minus 6 straight y space equals space 0
which is required differential equation.